PLANE    AND    SOLID 


Analytic   Geometry 


FREDERICK   H.  BAILEY,  A.M.  (Harvard) 

AND 

FREDERICK   S.  WOODS,  Ph.D.  (Gottingen) 


Assistant  Professors  of  Mathematics  in  thb 
Massachusetts  Institute  of  Technology 


Boston,  U.S.A.,  and  London 
GINN   &   COMPANY,   PUBLISHERS 

1897 


Copyright,  1897,  by 
F.  H.  BAILEY  AND  F.  S.  "WOODS 

ALL  RIGHTS  RESERVED 


PREFACE. 


EMS 
Lib. 

QA 


This  book  has  been  prepared  primarily  for  the  use  of  the 
students  in  the  Massachusetts  Institute  of  Technology,  but  it 
is  hoped  that  it  will  be  found  adapted  to  the  needs  of  other 
technical  schools  and  colleges. 

While  the  authors  have  restricted  themselves  to  subject- 
matter  properly  belonging  to  a  first  course,  they  have,  never- 
theless, endeavored  to  give  a  complete  and  rigorous  treatment 
of  all  questions  discussed.  The  memorizing  of  a  mass  of 
formulas  has  been  discouraged,  the  attention  of  the  student 
being  directed  rather  to  the  methods  employed.  At  the  end 
of  the  book  a  collection  of  formulas  has  been  made,  which,  it 
is  believed,  is  sufficient  for  all  needs  of  the  student. 

In  the  Plane  Geometry  considerable  space  has  been  devoted 
to  the  general  forms  of  the  equations.  In  particular,  the 
equation  of  the  circle  is  treated  in  its  most  general  form. 
The  conies  have  been  approached  from  tlieir  general  defini- 
tion, and  the  student  is  led  to  discuss  the  general  equation  of 
the  second  degree  in  which  the  xy  term  is  missing,  and  to 
find  the  equations  of  the  tangent,  the  normal,  and  the  polar 
for  the  equation  of  this  form.  By  this  method  of  procedure 
much  time  is  gained  when  the  special  equations  of  the  parab- 
ola, the  ellipse,  and  the  hyperbola  are  discussed,  and  it  is 
believed  that  the  student's  grasp  of  the  subject  is  thereby 
strengthened.     Finally,  the  general  equation  of  the  second 


IV  PREFACE. 

degree  has  been  studied  with  the  view  of  enabling  the  stu- 
dent to  determine  most  rapidly  and  easily  the  nature  and  the 
position  of  the  locus  of  any  equation  of  that  degree. 

In  the  Solid  Geometry,  besides  the  plane  and  the  straight 
line,  the  cylinders  and  the  surfaces  of  revolution  have  been 
noticed,  and  all  the  quadric  surfaces  have  been  studied  from 
the  simplest  forms  of  their  equations.  This  study  includes 
the  treatment  of  the  tangent,  the  polar,  and  the  diametral 
planes,  conjugate  diameters,  circular  sections,  and  rectilinear 
generators. 

The  examples  in  the  earlier  parts  of  both  the  Plane  and 
the  Solid  Geometry  are  mainly  numerical,  but  later  they  are 
largely  purely  geometrical.  They  have  been  arranged  at  the 
end  of  each  chapter,  that  the  teacher  may  take  them  up  in 
any  order  preferred.  Many  of  them  have  been  prepared 
especially  for  this  book,  and  for  the  others  the  authors  wish 
to  acknowledge  their  indebtedness  to  other  writers,  especially 
to  Professors  Eunkle,  C.  Smith,  and  Todhunter. 

In  conclusion,  the  authors  wish  to  thank  Mr.  H.  K.  Burrison, 
who  drew  the  diagrams  for  Chapters  II  and  IV  of  Part  II,  and 
other  friends  who  have  kindly  aided  them  by  suggestions. 

Massachusetts  Institute  of  Technology, 
February,  1897. 


OOIl^TENTS. 

PART  I. 
PLANE    ANALYTIC    GEOMETRY. 


CHAPTER  I. 

THE   POINT. 

ARTICLE  PAGE 

1.  Direction  of  a  Line  .........  1 

2.  Cartesian  Coordinates 2 

3.  Distance  between  Two  Points  and  Slope  of  Joining  Line         .  4 

4.  Point  of  Division      .........  7 

5.  Area  of  Triangle 9 

0.     Area  of  Polygon 11 

7.  Polar  Coordinates 13 

8.  Distance  between  Two  Points  and  Slope  of  Joining  Line,  in 

Polar  Coordinates .  14 

9.  Area  of  Triangle  in  Polar  Coordinates 15 

10.     Relation  between  Cartesian  and  Polar  Coordinates          .         .  16 

Examples 17 


CHAPTER  IL 


IiOCI. 


11.  Relation  between  Equation  and  Locus 

12.  Equation  of  the  First  Degree    . 

13.  Equations  of  Degree  Higher  tlian  the  First 

14.  Transcendental  Equations 

15.  Equations  in  Polar  Coordinates 

16.  Locus  Defined  by  Geometric  Property 

17.  Intersection  of  Loci  (Case  I)     . 

18.  Intersection  of  Loci  (Case  II)  . 

19.  Limiting  Cases  of  Intersection 
Examples 


22 
23 
24 
28 
30 
32 
32 
35 
38 
40 


vi  CONTENTS. 


CHAPTER  III. 

THE    STRAIGHT    LINE!. 
ABTICLK  PAGE 

20 45 

21.  Equation  of  the  Straight  Line  in  terms  of  its  Slope  and  Liter- 

cept  on  the  Axis  of  y 45 

22.  Equation  of  the  Straight  Line  in  terms  of  its  Intercepts  on  the 

Axes 47 

23.  Equation  of  the  Straight  Line  in  terms  of  its  Normal  Distance 

from  the  Origin  and  the  Angle  the  Normal  makes  with  the 
Axis  of  X       .         .         .         .         .         .         .         .         .         .49 

24.  Any  Equation  of  the  First  Degi'ee  represents  a  Straight  Line  .  50 

25.  Lemma     ...........  52 

26.  Determination  of  the  Parameters  of  any  Straight  Line     .         .  52 

27.  Equations  connecting  the  Parameters  of  the  Straight  Line       .  55 

28.  Equations  of  the  Straight  Lme  in  Oblique  Coordinates     .         .  56 

29.  Lemma     ...........  58 

30.  Equations  connecting  the  Parameters  of  the  Straight  Line  in 

Oblique  Coordinates 59 

31.  Equation  of  the  Straight  Line  in  Polar  Coordinates          .         .  60 

32.  Distance  of  a  Point  from  a  Straight  Line  .         .         .         .60 

33.  Equation  of  a  Line  through  any  Given  Point  with  a  Given  Slope  63 

34.  Equation  of  a  Line  through  Two  Given  Points  .         .         .64 

35.  Angle  between  Two  Lines 65 

36.  Equation  of  a  Line  through  a  Given  Point  Perpendicular  to  a 

Given  Line 68 

37.  Problem 69 

38.  Equation  of  a  Line  through  the  Intersection  of  Two  Given  Lines  70 

39.  Equations  of  the  Bisectors  of  the  Angles  between  Two  Lines  .  72 

40.  Equations    of    Higher  Degree    than  the    First  representing 

Straight  Lines 74 

41.  Condition  that  the  Equation  of  the  Second  Degree  represents 

Two  Straight  Lines 75 

Examples 77 

CHAPTER  IV. 
TRANSFORMATION  OP  COORDINATES. 

42.  Meaning  of  Transformation  of  Coordinates      .         .         .         .86 

43.  Translation  of  Origin 87 


CONTENTS.  Vll 

ABTICLE  PAGE 

44.  notation  of  Rectangular  Axes 88 

45.  Transforuiatiou  from  a  Set  of  Rectangular  Axes  to  a  Set  of 

Oblique  Axes  having  the  Same  Origin  ....  89 
40.     Transformation  from  one  Set  of  Oblique  Axes  to  a  new  Set 

of  Oblique  Axes  with  the  same  Origin         ....  90 

47.     Double  Transformation  of  Coordinates 92 

48 93 

Examples        ..........  93 

CHAPTER  V. 
THE   CIRCIjE. 

49.  Definition  and  Equation  of  the  Circle 97 

5(4.     General  Form  of  the  Equation  of  the  Circle   ....  98 

51.  Determination  of  the  Centre  and  the  Radius  of  a  Circle  from 

its  Equation 99 

52.  Determination  of  the  Ecjuation  of  a  Circle  Passing  through 

Three  Given  Points 100 

53.  Determination  of  the  Tangent  to  a  Circle  when  its  Slope  is 

Given 101 

54.  Equation  of  a  Tangent  to  a  Circle  when  the  Point  of  Contact 

is  Given 108 

55.  Normal 105 

50.  Subtangent  and  Subnormal 1()(; 

57.  Length  of  a  Tangent  Drawn  from  a  Point  to  a  Circle     .         .  107 

58.  Chord  of  Contact 1(»H 

59.  Definition  and  Equation  of  the  Polar 110 

00.  Other  Properties  of  Poles  and  Polars 113 

01.  Diameter 115 

62.     Circle  through  the  Points  of  Intersection  of  'JVo  Circles         .  118 

03.  Radical  Axis 119 

04.  Radical  Centre 121 

65.     Polar  E(iuation  of  the  Circle 122 

Examples        ..........  123 

CHAPTER   VI. 
THE    CONIC    SECTIONS. 

60.  Definition  and  Equation 131 

67.     The  Parabola  [e  =  1]      ........  133 


Vlll 


CONTENTS. 


ARTICLE  PAGE 

08.     The  Ellipse  [e  <  1] 135 

69.  The  Hyperbola  [e  >  1] 140 

70.  Generalization 144 

71.  Limiting  Cases  of  Conic  Sections 149 

72.  Recapitulation 150 

73.  Sections  of  a  Cone 151 

74.  Polar  Equation  of  the  Conic  Section,  the  Focus  being  the  Pole  157 
Examples 157 


CHAPTER  VII. 
TANGENT,  NORMAL,  AND   POIjAR. 

75 161 

76.  Tangent 161 

77.  Normal 165 

78.  Pole  and  Polar 106 

79.  To  find  the  Tangents  from  a  Point  not  on  the  Conic  Section  169 

80.  Properties  of  Poles  and  Polars 171 

Examples 174 


CHAPTER  VIII. 


THE3  PARABOLA:    y^  =  4px 


Constructions 

Angular  Properties  of  Tangent  and  Normal 
Perpendicular  Tangents  meet  on  the  Directrix 
Perpendicular  to  Focal  Chord 
Perpendicular  from  Focus  to  Tangent    . 
Diameter         ...... 

Properties  of  the  Diameter 
Parabola  with  Focus  at  the  Origin 
Parabola  referred    to  a  Diameter  and    the 
Tangent  as  Axes  .... 

Parabola  in  Polar  Coordinates 
Examples 


Corresponding 


177 
177 
179 
181 
182 
183 
186 
188 
189 

191 
193 
193 


CONTENTS. 


IX 


CHAPTER  IX. 


THE    EHiIilPSE!: 


X.-      y2 


AKTICLE 
92.       . 

93. 
94. 
95. 


Parallel  to 


To  find  the  Foci  when  the  Axes  are  givei 

Focal  Distances 

Sum  of  Focal  Distances 

96.  Constructions 

97.  Angle  between  Tangent  and  Focal  Radii 

98.  Point  of  Intersection  of  a  Pair  of  Perpendicular  Tangents 

99.  Perpendicular  from  the  J"ocus  upon  any  Tangent 

100.  Equation  of  a  Diameter  .... 

101.  Conjugate  Diameters 

102.  The  Tangent  at  the  Extremity  of  any  Diameter  is 

the  Conjugate  Diameter      .... 

103.  Supplemental  Chords 

104.  Given  the  Extremity  of  any  Diameter,  to  find  the  Extremities 

of  the  Conjugate  Diameter 

105.  Lengths  of  Conjugate  Diameters    . 

106.  Angle  between  Conjugate  Diameters 

107.  Parallelogram  on  Conjugate  Diameters 

108.  Auxiliary  Circle      .... 

109.  Eccentric  Angle      .... 

110.  Eccentric  Angles  and  Conjugate  Diameters    . 

111.  Ellipse  referred  to  Conjugate  Diameters  as  Axes 

112.  Polar  Equation  of  the  Ellipse 
Examples 


PAGE 
199 

199 
200 
201 
202 
202 
203 
205 
206 
208 

209 
210 

212 
213 
214 
215 
216 
217 
218 
220 
222 
223 


CHAPTER  X. 


THE   HYPERBOIiA: 


113.  Conjugate  and  Equilateral  Hyperbolas  . 

114.  To  find  the  Foci  when  the  Axes  are  given 

115.  Focal  Distances 

116.  Difference  of  Focal  Distances 

117.  Constructions.         ..... 

118.  Angle  between  Tangent  and  Focal  Radii 

119.  Equation  of  a  Diameter  .... 


229 
231 
232 
233 
233 
234 
2.30 


X  CONTENTS. 

ARTICLE  PAGE 

120.  Conjugate  Diameters 237 

121.  Propositions  on  Conjugate  Diameters 238 

122-126.     Asymptotes 240 

127.  Equation  of  an  Hyperbola  referred  to  its  Asymptotes  as  Axes 

of  Coordinates 244 

128.  Polar  Equation  of  the  Hyperbola 246 

Examples 246 


CHAPTER  XL 
THE   GENERAIj   E5QUATION   OF   THE   SECOND   DEGREE 

129.  Statement  of  Problem 251 

130.  To  Remove  Terms  of  the  First  Degree 252 

131.  Case  I.     AB-H2>0 253 

132.  Case  II.     AB-H2<0 258 

133.  Casein.     AB  -  H-^  =  0 260 

134.  Summary 262 

135.  Rule  for  Handling  Numerical  Equations         ....  263 

136.  Oblique  Coordinates 267 

137.  Conic  through  Given  Points 268 

Examples        ..........  271 


PAKT  II. 
SOLID   ANALYTIC    GEOMETRY. 


CHAPTER  I. 

THE  POINT. 

1. 

2. 
3. 
4. 
5. 
6. 

Rectangular  Coordinates  in  Space  . 

Polar  Coordinates  in  Space     .... 

Distance  between  Two  Points 

Point  of  Division 

Direction  Cosines 

Projection 

.     273 
.     275 

.     277 
.     278 
.     279 
.     281 

CONTENTS.  Xi 

ARTICLE  PAGE 

7.  Angle  between  Two  Lines 282 

8.  Translation  of  Origin 283 

9.  Transformation  of  Coordinates  from  One  Set  of  Rectangular 

Axes  to  a  New  Set  of  Rectangular  Axes  having  the  Same 

Origin 284 

10 286 

11.     Relation  between  Rectangular  and  Polar  Coordinates    .        .  286 

Examples 288 


CHAPTER  II. 


INTERPRETATION    OP    EQUATIONS. 


12.  Single  Equation 

13.  Two  or  More  Equations  . 

14.  Cylinders 

15.  Surfaces  of  Revolution    . 
Examples 


290 
292 
293 
294 

298 


CHAPTER  III. 
THE    PLANE   AND    THE    STRAIGHT    IiINB. 

16.  Any  Equation  of  the  First  Degree  Represents  a  Plane  .         .  301 

17.  Normal  Equation  of  the  Plane        .         .         .         .         .         .  302 

18.  Equation  of  a  Plane  in  Terms  of  its  Intercepts       .         .         .  303 

19.  To  Find  the  Equation  of  a  Plane  which  Passes  through  Three 

Given  Points 304 

20.  Angle  between  Two  Planes 304 

21.  Perpendicular  Distance  from  a  Given  Point  to  a  Given  Plane  305 

22.  Any  Two  Simultaneous  Equations  of  the  First  Degree  Repre- 

sent a  Straight  Line 306 

23.  Equations  of  a  Straight  Line  in  Terms  of  its  Direction  Cosines 

and  Any  Known  Point  upon  it 306 

24.  To  Place  the  General  Equations  of  the  Straight  Line  in  the 

Form  [14]  .    ' 308 

25.  Equations  of  a  Straight  Line  Passing  through  Two  Given  Points  309 

26.  Plane  through  a  Given  Line  and  Subject  to  One  Other  Condition  310 
Examples 311 


XU  CONTENTS. 

CHAPTER  IV. 

EQUATIONS    OF    THE    SECOND    DEGREE. 

AKTICLE  PAGE 

27.  The  Cones  :  -,  ±  f^  ±  ^  =  0 316 

a^      b^      c^ 

28.  The  Central  Quadrics :   -,  ±  ^,  ±  ^  =  1         .         .         .         .320 

a-      b-      c2 

X^  y'2  -^1 

29.  Relation  between  the  Cone  -:  4-  7-, ;  =  0   and  the  Hyper- 

a?      b-      c^ 

boloid  ^  +  ^  -  ^  =  1 328 

a^       b-      c^ 

30.  The  Unparted  Hyperboloid  is  a  Ruled  Surface        .        .         .     329 

31.  The  Paraboloids :-,  ±  ^"  =  4pz 330 

a-      b^ 

32 330 

33.  Tangent  Plane 332 

34.  Normal  Line 333 

35.  Polar  Plane 334 

36.  Diametral  Plane 335 

37.  Diameters 337 

38.  Auxiliary  Line  of  the  Ellipsoid 339 

39.  Parallel  Sections 340 

40.  Circular  Sections 341 

41.  Umbilic 343 

Examples 345 

FORMULAS 351 

ANSWERS 361 


PAET   I. 

PLANE    ANALYTIC    GEOMETRY. 
CHAPTER  I. 

THE    POINT. 

1.     Direction  of  a  Line. 

The  line  AB,  A B,  joining  the  points  A  and 

B,  may  be  regarded  as  having  two  directions,  according  as 
it  is  drawn  from  A  to  B  or  from  B  to  A.  To  distingnisli 
between  these  two  directions  we  will  adopt  the  following 
notation  : 

If  the  line  is  drawn  from  A  to  B,  if  shall  he  called  AB ;  if  it 
is  drawn  from  B  to  A,  it  shall  be  called  BA. 

The  first  letter,  then,  denotes  the  beginning,  or  the  origin, 
of  the  line,  and  the  last  letter  denotes  the  end,  or  the  term, 
of  the  line. 

If,  now,  A,  B,  and  C  are  any  three  points  on  the  same 
straight  line, 

A B C  A C  B 

B A C         C A B 

Fig.  1. 

it  follows  that  AB  +  BC  =  AC  (1) 

For  the  resnlt,  in  both  distance  and  direction,  of  going  from 
A  to  B  and  then  from  B  to  C  is  the  same  as  the  resnlt  of 
going  from  A  to  C,  as  shown  in  Fig.  1,  which  illustrates  some 
of  the  cases  which  may  occur. 


2  PLANE   ANALYTIC   GEOMETRY. 

If,  however,  the  third  point  C  coincides  with  A,  (1)  becomes 

AB+  BA=^  AA=0, 
whence  BA  =  —  AB. 

That  is,  if  one  direction  on  a  straight  line  is  jiositive,  the  oppo- 
site direction  is  negative,  so  that  AB  and  BA  are  each  the 
negative  of  the  other. 

From  (1)  we  may  derive  the  formula, 

BC=AC-AB;  (2) 

i.e.,  if  three  j^oints  lie  iij/on  the  same  straight  line,  the  line 
joining  any  one  of  them  to  a  second  is  equal  to  the  line  join- 
ing the  third  to  the  second  decreased  by  the  line  joining  the 
third  to  the  first. 


2.     Cartesian  Coordinates. 

Since  the  first  element  of  geometrical  discussion  is  the 
point,  let  us  now  find  a  method  by  which  we  can  determine 
the  relative  position  of  a  point  in  a  plane  with  respect  to 
some  fixed  positions  in  the  plane.  We  will  first  take  the 
method  invented  by  Descartes. 

Y 


Fig.  2. 


THE   POINT.  d 

Let  XX'  and  YY'  be  two  straight  lines  intersecting  at  0. 
Distances  from  YY',  measured  parallel  to  XX',  shall  be  de- 
noted by  X,  X  being  positive,  if  measured  in  the  direction 
OX,  and  negative,  if  measured  in  the  direction  OX'.  Dis- 
tances from  XX',  measured  parallel  to  YY',  shall  be  denoted 
by  y,  y  being  positive,  if  measured  in  the  direction  OY,  and 
negative,  if  measured  in  the  direction  OY'. 

Then  there  can  be  but  one  value  of  x  and  one  value  of  y 
corresponding  to  any  point  in  the  plane.  For  tlirougli  the 
point  we  can  draAV  only  one  line  parallel  to  YY',  and  only 
one  line  parallel  to  XX',  and  the  respective  distances  of 
these  lines  from  YY'  and  XX',  measured  according  to  the 
rule  laid  down  above,  will  be  the  respective  values  of  x  and 
y  for  the  point. 

On  the  other  hand,  if  any  simultaneous  values  are  given 
to  X  and  y,  these  values  determine  the  position  of  a  single 
point  in  the  plane.  For  example,  let  x  ^=  a  and  y  ^  b,  wliere 
a  and  b  are  positive  quantities.  Then  the  point  is  in  a  line 
parallel  to  YY',  and  a  units  to  the  right  of  it,  and  also  in  a 
line  parallel  to  XX',  and  b  units  above  it.  These  two  lines 
meet  in  a  single  point,  Pi,  as  shown  in  Fig.  2.  Or,  again,  let 
x^  —  a,  y  =  — b;  then  the  point  is  in  a  line  parallel  to  YY', 
a  units  to  the  left  of  YY',  and  in  a  line  parallel  to  XX', 
b  units  below  XX',  and  the  point  P2  is  determined,  as  sliown 
in  Fig.  2.  Thus  we  see  that,  given  simultaneous  values  of 
X  and  y,  we  can  find  one  and  only  one  point  corresponding 
to  them.  Practically,  the  position  of  this  point  may  be  found, 
or,  as  we  usually  say,  the  point  may  be  j^lotted,  by  first  laying 
off  X  along  XX',  and  then  measuring  y  on  a  line  parallel 
to  YY';  for  example,  in  Fig.  2,  by  laying  off  OMi,  and 
then  Ml  Pi. 

The  distances  x  and  y  are  called  respectively  the  abscissa 
and  the  ordinate  of  the  point ;  or,  taken  together,  they  are 
called  the  coordinates  of  the  point.  The  lines  XX'  and  YY' 
are  the  axes  of  abscissas  and  ordinates  respectively,  or  the 


4  PLANE    ANALYTIC    GEOMETRY. 

coordinate  axes,  and  their  point  of  meeting  is  called  the 
origin. 

Tlie  axes  may  meet  at  any  angle,  and  we  have  rectangular 
coordinates,  or  oblique  coordinates,  according  as  the  axes  are 
or  are  not  perpendicular  to  each  other.  Both  classes  of  coor- 
dinates, however,  fall  under  the  class  of  Cartesian  coordinates, 
named  from  the  inventor  Descartes,  the  Latin  form  of  whose 
name  is  Cartesius.  If  no  angle  between  the  axes  is  specified, 
we  shall  assume  it  to  be  a  right  angle. 

Instead  of  writing  x  =  —  a,  y  =  —  b,  we  shall  always  write 
P  ( —  a,  —  b)  or  ( —  a,  —  b),  for  the  sake  of  brevity,  the  abscissa 
always  being  written  first  and  separated  from  the  ordinate 
by  a  comma. 

3.    Distance  between  Two  Points  and  Slope  of  Joining  Line. 

Now  that  we  can  determine  the  position  of  a  point  in  the 
plane,  let  us  see  how  we  can  use  the  coordinates  of  points 
in  a  geometrical  problem.  The  first  problem  is  that  of  find- 
ing the  distance  between  two  points,  which  shall  be  Pi  (xi,  yi) 
and  P2  (x2,  y2). 


Fig.  3. 


THE    POINT.  O 

From  P2  and  Pi  draw  lines  respectively  parallel  to  XX'  and 
YY',  meeting  at  R. 

Then,  OMi=:x„   MiPi  =  yi, 

OM,=  Xo,   M,P.  =  yo. 

By  (2)  §  1,  P,R=M,Mi  =  OMi-OMo  ^Xj-x., 

RP^=  MiPi- MiR=  MiPi- MoP,  =  yi-y,. 

If  we  denote  the  distance   PiPo  by  d,  we  have,  from  the 
right  triangle  PiPoR, 

whence,  by  substituting  the  values  of  PoR  and  RPi, 


d-V(Xi-x.)^'  +  (y:-y,)^.  [1] 

If  6  denotes  the  angle  made  with  the  axis  XX'  by  the  line 

RPi              .     Yi  — ya 
Pi  P2,  tan  0  =  H-B»  or  tan  0  = -• 

r  2r»  Xi  —  Xo 

Tan  $  is  called  the  slope  of  the  line,  and  is  usually  denoted 
by  m. 

Since  6  and  m  are  quantities  constantly  occurring  in  the 
following  work,  it  is  important  to  notice  carefully  how  6  is 
to  be  measured.  It  is  always  the  angle  above  the  axis  of  x 
and  to  the  right  of  the  line  P1P2,  the  line  being  produced,  if 
necessary,  so  as  to  cross  XX'.  Its  angular  magnitude  is,  there- 
fore, always  between  0°  and  180°.  Consequently,  m  may 
have  any  value  whatever,  and  is  positive  when  9  is  an  acute 
angle,  and  negative  when  0  is  obtuse.  Conversely,  when  m 
is  known,  6  is  determined  without  ambiguity. 

In  Fig.  3,  from  which  we  have  derived  formulas  [1]  and  [2], 
both  Pi  and  P2  are  in  the  first  quadrant,  but  the  reasoning 
will  apply  to  any  position  of  the  two  points.  For  example, 
it  applies  to  the  following  diagrams,  Figs.  4  and  5,  without 
a  single  change  of  wording. 


PLANE   ANALYTIC    GEOMETRY. 


P, 


IR 


Y 

Fig.  4. 


O  M, 


Y 

Fig.  5. 


Ex.     Find  m  and  d  for  line  joining  tlie  points  (2,  3)  and  {—  1,  2). 


d  =  V[2  -  (-  1)]2  +  [3  —  2]2  =  VlO, 


3  —  2 
2-(-l) 


If  the  axes  make  an  angle  w  with  each  other,  we  shall  have 
to  find  new  formulas  for  d  and  m.  Accordingly,  let  XX'  and 
YY',  in  Fig.  6,  meet  so  that  Z  XOY  =  (o. 


Fig.  6. 


THE    POINT.  7 

Pi  (xi,  yi)  and  P.2  (xo,  72)  are  the  two  points.  Draw  PiR 
parallel  to  YY',  PoR  parallel  to  XX',  PjS  perpendicular  to 
PoR,  and  P.M.,  parallel  to  YY'.     Then, 

0M.,  =  X2,    M,P.,  =  y.3. 

.-.  P.,R=  M,Mi  =  0Mi-0Mo  =  xr-X2, 
RP^=M,Pi-MiR  =  yi-y.,. 

In  the  triangle  PiRP,>, 


P^P,=zAyp,R-+RP,'-2  P.,R.RPi  cos  Z  PaRPi- 

As  cos  ZP2RPi^cos   (180°  —  co)  =  —  cos  w,  we  have  by 
substitution 

d  =  V(xi  —  X.,)"  +  (yi  —  yo)-  +  2  (xi  —  Xo)  (yi  —  y^)  cos  u>.  (1) 

Now,      RS=RPi  cos  w=(yi — yo)  cos  w, 

so  that      PoS  =  PoR  +  RS  =  (Xi  —  x,)  +  (yi  —  y2)  cos  w  ; 
and  SPi  ^  RPi  sin  w  =  (y,  — y.,)  sin  w. 

SP 
But  m  =  tan  $  =  p— '  ; 

r  2^ 

. (y,  — ya)  sin  o)  ^ 


(xi  — x2)  +  (yi  — yo)  cos 


It  is  evident  that,  if  w  =  90°,  (1)  and  (2)  reduce  respec- 
tively to  [1]  and  [2],  as  we  should  expect. 

4.     Point  of  Division. 

The  next  problem  that  naturally  occurs  is  that  of  finding 
a  point  which  divides  a  line  in  a  given  ratio.  We  will  find, 
then,  the  coordinates  of  the  point  P,  on  the  line  joining  Pi 
(xi,  yi)  and  P,  (x.,  y.,),  such  that   P,P  :    PP^^  I,  :  L 


PLANE   ANALYTIC   GEOMETRY. 
Y 


HR 


Fig.  7. 

Draw  PiMi,  PM,  cand  P2M2  all  parallel  to  YY',  and  PiR  aud 
PS  parallel  to  XX'. 

Then,  P^R=  MiM=:  OM  —  OMi  =  x— x^, 

PS  =MM,  =  OM,-OM  =Xo-x, 
RP=MP  -MR  =MP  -MiPi=y-yi, 
SP,=  M2Po-MoS=M,Po-  MP  =  y,-y. 

The  triangles  P^RPand  PSP2  are  similar,  since  their  sides 
are  respectively  parallel;  therefore  their  homologous  sides 
are  proportional,  and 

PiP_  PiR^RP 

PP2~      PS~SPo" 

Substituting  the  respective  values  we  have  found,  we  get 


i, 


whence 


x== 


X  —  xi  ^  y 
Xo  —  X        y 

llXo    +    loXl 


I1+I2 


y 

and  y 


Iiy2  +  l2yi 
I1  +  I2 


[3] 


Ex.     Find  the  coordinates  of  a  point  dividing  tlie  line  joining  (1,  4)  to 
(3,  —  2)  in  the  ratio  2  :  3. 


THE    POINT.  y 

If  we  let  (1,  4)  be  (xi,  yi),  aud  (3,  —  2)  be  (xo,  yc),  then  li  =  2,  and 
I2  =  3,  and  the  coordinates  of  the  required  points  are 

^  2(3)  +  3(1)  ^  9  2  (-  2)  +  3(4)  ^  8 

2  +  3  5'       ^  2  +  3  5' 

If  the  point  bisects  the  line  P1P2,  Ii  =  l2j  and  the  formulas 
become 

If  the  point  is  one  of  external  division,  the  lines  PiP  and 
PP2  will  have  opposite  directions,  and  hence  their  ratio  must 
be  negative.  We  can  make  it  negative  by  putting  a  minus 
sign  before  I2,  in  which  case  our  formulas  become 

^  _  liX2-i2Xi       _  liy^  -  l2yi  r-n 

ii     12  11     "2 

As  our  proof  has  depended  entirely  upon  the  properties 
of  similar  triangles,  it  is  evident  that  it  would  be  exactly  the 
same  for  oblique  coordinates,  so  that  our  resulting  formulas, 
[3],  [4],  [5],  are  true  for  oblique  coordinates. 

And  it  may  be  noted  here,  that,  if  no  use  is  made  of  the 
right  angle  between  the  coordinate  axes,  the  derived  formula 
will  be  a  general  formula,  holding  for  all  cases  of  Cartesian 
coordinates,  even  though  in  the  figure  used  XX'  is  perpendic- 
ular to  YY'. 

5.  Area  of  Triangle  in  terms  of  the  Coordinates  of  the 
Vertices. 

Let  the  vertices  of  the  triangle  be  Pj  (xj,  yi),  P2  (xo,  y2)>  and 
P3  (X3,  ys)- 

Draw  PiN  parallel  to  XX',  and  P3M  and  P.N  parallel  to  YY', 
thus  forming  the  right  triangles  P1P3M  and  P1P2N,  and  the 
trapezoid  MP3P2N. 

Then  the  area  of  PiP2P3  =  area  of  P1P3M  -|-  area  of  M  PgPaN 
—  area  of  PiPoN. 


10 


PLANE  ANALYTIC   GEOMETRY. 
Y 


x'   ^'    / 

\ 

0           1M2 

\y^-^ 

M3 

1 
1 

p,            1^ 

/I 

yi' 


Y 

Fig.  s. 

Area  of  P1P3M  =  iPiM.MP3, 

''       "  MPsPoN^iMN  (MP3+NP2), 

u       u  PiP,N  =  iPiN.NP2. 

Now,         PiM  =  MiM3  =  OM3  — 0M,.  =  X3  — xi, 

MP3=M3P3-  M3M  =  M3P3-MiPi  =  y 
MN  =M3Mo  =  OM2-OM3  =  Xo-x3, 
N P,  =  M,P, -  MoN  =  M,Po  —  MiPi  =  yo - yi, 
P,N=  MiM2  =  0Mo  — OMi  =  Xo— xi; 

.•.by  substitution,  the  area  of  PiP2P3  = 

h  (^3  —  xi)  (ys  —  yi)  +  h  (^2  —  X3)  (ys  —  yi  +  y2  —  yO 

—  h  (x2  — xi)  (y2  — yi), 

whence  we  get,  in  reduced  form,  area  of  PiP2P3  = 

¥{(xiy2  — X2yi)  +  (xoy3  — X3y2)  +  (xsYi  — Xiy3)}.        [6] 

The  following  is  an  easy  method  of  solving  numerical  ex- 
amples in  accordance  with  the  above  formula  :  Write 
xj  yi  down  the  coordinates  of  the  successive  vertices  in  a 
Xo  y2  column,  repeating  at  the  bottom  of  the  column  the 
X3  ya  coordinates  of  the  first  vertex  ;  multiply  each  x  by 
Xi  yi     the  y  in  the  line  below  it,  prefixing  the  positive  sign. 


2, 

3 

1, 

5 

-1, 

—  2 

THE   POINT.  11 

and  each  y  by  the  x  in  the  line  below  it,  prefixing  the 
negative  sign,  and  take  one  half  the  algebraic  sum  of  all  the 
terms  thus  formed  ;  the  result  is 

I  {xiy2  —  Xoyi  +  xoys  —  xsya  +  xsyi  —  xiya} , 

which  we  have  proved  to  be  the  expression  for  the  area. 
It  is  to  be  noted  that  this  last  is  only  a  working  rule, 
giving  a  true  result,  and  is  in  no  sense  a  proof  of  the 
formula. 

Ex.     Find  the  area  of  the  triangle  of  which  the  vertices 
are  the  points  (2,  3),  (1,  5),  and  {—  1,  —  2). 

Area     =  i  \  (2)(5)  -  (1)(3)  +  (1)(-  2)  -  (-  1)(5) 
2,      3  +(-l)(3)-(2)(-2)|=-V-. 

If  the  expression  for  the  area  of  any  triangle  comes  out 
negative,  the  absolute  value  of  the  result  is  the  area,  and  the 
negative  sign  shows  that  we  have  lettered  the  vertices  so  that 
in  going  from  Pj  to  Pa  and  then  from  Pg  to  Pg,  we  have  the 
triangle  on  our  right,  while,  as  we  have  lettered  Fig.  8,  we 
have  it  on  our  left.  That  such  is  the  meaning  of  the  negative 
sign  may  be  shown  by  interchanging  P2(x2,  y^)  and  P3(x3,  yg) 
in  Fig.  8,  and  finding  the  formula  for  the  area  of  the  triangle, 
the  result  differing  from  that  already  found  in  the  negative 
sign  only. 

Note.  Fig.  8  is  necessarily  drawn  for  some  chosen  position  of  the 
triangle,  but  the  proof  as  given  will  apply  to  all  cases,  if  the  vertices  be 
appropriately  lettered.  It  is  suggested  that  the  student  make  other 
figures  from  which  to  derive  the  formula,  as  in  §  3. 

6.  Area  of  a  Polygon  in  terms  of  the  Coordinates  of  its 
Vertices. 

We  may,  of  course,  find  the  area  of  any  polygon  of  four  or 
more  sides,  by  dividing  it  into  triangles  by  diagonals,  and 
taking  the  sum  of  the  areas  of  the  triangles  thus  formed. 


12 


PLANE   ANALYTIC   GEOMETRY. 


But  we  can  deduce  a  formula  for  the  area  of  any  polygon, 
exactly  similar  to  that  for  the  area  of  a  triangle,  and  giving 
the  same  working  rule.  This  causes  a  great  saving  of 
time  over  the  method  first  suggested. 

We  will  illustrate  the  method  by  finding  the  formula  for 
the  area  of  the  quadrilateral  P1P2P3P4,  the  coordinates  of  the 
vertices  being  respectively  (xj,  yj),  (x^,  y^),  (xs,  ys),  and  (X4,  y^). 


Let  P  (x,  y)  be  any  point  within  the  quadrilateral,  and  draw 
the  lines  PPi,  PPg,  PP3,  and  PP4,  forming  the  triangles  PP1P2, 
PP2P3,  PP3P4,    and  PP^Pi.     By  [6]  §5, 

Area  of  PPiPo  =  i{xyi  —  Xjy  +  Xiy,  — xsyi  +  x.y  —  xyz}, 

•     "  "  PP2P3  =  |{xy2  — x.2y  +  x2y3  — X3y2+X3y  — xys}, 

"  "  PP3P4  =  i{xy3  — xsy  +  xgyi— X4y3+X4y  — xy4},    • 

"  "  PP4Pi  =  i{xy4  — X4y  +  X4yi  — Xiy^  +  xiy  — xyj. 

Adding  and  reducing,  we  get 
Area  of       Pi  P2  P3  P4  =  i  {xiy2  —  X2y  1  +  Xays  —  Xsy-  +  xsy* 


—  X4y3+  X4yi  — Xiy4}. 


(1) 


THE    POINT.  13 

Trying  a  working  rule  like  that  for  the  triangle,  we  get 

xi  yi  Area  of     PiP2P3P4=  i{xiy2  —  Xoyi  +  x^ys  —  X3y2  +  X3y4 
X2  y2  —  X4y3  + x^yi  — x,y4}, 

X3  y,3  a  true  result  agreeing  with  (1). 

X4  y4         This  may  be  extended  to  a  polygon  of  any  number 

Xi  yi  of  sides,  the  resulting  fornuda  being,  if  the  vertices  are 

(xi,  yi),  (X2,  yo)  .  .  .  .  (x„,  y„), 
h  {(xiy2  —  xoyi)  +  (xoya  —  xsy-)  +  •  •  •  +  (x„yi  —  xiy„)} .       (2) 

7.     Polar  Coordinates. 

So  far,  we  have  determined  the  position  of  the  point  in  the 
plane  by  two  distances,  i.e.,  by  x  and  y.  We  may,  however, 
determine  the  position  of  the  point  by  a  distance  and  a  direc- 
tion, as  follows : 


Fig.  10. 

Let  0,  called  tlie  origin  or  pole,  be  a  fixed  point  in  the 
plane,  and  OM,  called  the  initial  line,  be  a  line  of  fixed 
direction  in  the  plane. 

Take  P  any  point  in  the  plane,  and  draw  OP.  Denote  OP 
by  r,  and  the  angle  MOP  by  ^  ;  then  r  and  6  will  be  called 
the  polar  coordinates  of  the  point,  and,  when  given,  will  com- 
pletely determine  the  point. 


14 


PLANE   ANALYTIC    GEOMETRY. 


For  example,  let  r  =  2  and  $  =  15° ;  then  the  point  is  on  the 
circumference  of  a  circle  of  radius  2,  the  centre  of  which  is 
at  0,  and  on  a  straight  line  drawn  from  0,  making  ^  MOP  = 
15°.  These  two  lines  meet  in  one,  and  only  one,  point,  i.e., 
P,  the  position  of  which  is  completely  determined. 

For  brevity,  we  may  write  P  (2,  15°)  to  designate  this 
point,  for  which  r  =  2  and  6  =  15°.  The  point  P  is  j^ioUed  by 
laying  off  the  angle  15°,  OM  being  taken  as  the  initial  line, 
and  measuring  from  0  a  distance  2  along  the  terminal  line. 

The  angle  6  may  have  any  magnitude,  and  may  be  either 
positive  or  negative,  OM  being  regarded  always  as  the  initial 
line  of  the  angle,  as  implied  by  its  name.  Generally,  r  will 
be  positive,  but  we  may  give  r  a  negative  value,  if  we  make 
the  following  convention :  After  the  angle  $  is  constructed,  r 
shall  be  positive  if  the  point  is  on  the  terminal  line  of  6,  and 
negative  if  the  point  is  on  the  extension  of  the  terminal  line 
of  6,  the  terminal  line  being  extended  backward  through  the 
origin.  Thus  the  coordinates  of  Pi,  Fig.  10,  may  be  (2, 195°), 
or  (2,-165°),  or  (-2,  15°),  or  (-2,  —345°). 


8.     Distance  between   Two  Points  and  Slope  of  Joining 
Line,  in  Polar  Coordinates. 


/. 


,-^ 


Fig.  11. 


Let  Pi  (ri,  ^i)  and  P2  (r2,  O^)  be  the  two  points,  and  draw 
OPi  and  OP2. 


THE   POINT. 


15 


Then  in  the  triangle  PoOPi,  Z  PoOPi  =  ^i  — ^2,  so  that,  by 
Trigonometry, 

p;p;'=op;'+op"/— 20P1.OP2COS  (^1—^2); 

.".  if  the  length  of  P1P2  is  denoted  by  d, 


d  =  Vri^  +  r2-  -  2rir2  cos  {6^  —  6^). 


(1) 


Draw  PjRjand  P2R2  perpendicular  to  OM,  and  P2S  perpen- 
dicular to  PiRi-  Then,  if  we  detine  the  slope  as  the  tangent 
of   the   angle   the   line  makes    with    OM,  it  is  evident  that 

SPi 

^=p7s- 

But  SP,=  RiPi—  R2P2  =  ''i  sin  B^—r^  sin  6^, 

PoS^RsRi^ORi— 0R2=ri  cos  ^1  — r2  cos  ^2; 


ri  sin  c^i  —  r.i  sm  6^2 

.-.  m  = —• 

ri  cos  Bi  —  To  cos  d-i 

9.     Area  of  Triangle  in  Polar  Coordinates. 


(2) 


Fig.  12. 


Let  the  vertices  of  the  triangle  be  Pi  (ri,  ^1),  P2  (ro,  B^,  and 
P3  (""3,  ^3).  Connecting  the  origin  with  the  vertices  of  the 
triangle,  we  shall  form  three  triangles,  P1OP2,  PaOPj,  PsOPi, 


16 


PLANE   ANALYTIC   GEOMETRY. 


such  that  area  of  PiPoPs^area  of  PiOPa  +  areaof  P2OP; 
area  of  P3OP1. 
By  Trigonometry, 

Area  of  PiO Po  =  ^  nr.  sin  (0,  —  O^), 

"     "  P,OP3  =  ir,r3  sin  (^2-^3), 

"     "  P30Pi  =  ir3ri  sin  (^1-^3); 

.-.Area  of  P1P2P3  =  H^i^^  ^in  {O^  —  e^) 

+  x^x^  sin  {62  —  O3)  +  XsXi  sin  (^3  —  Oi)} . 

Note.  It  is  better  to  regard  formulas  (1)  and  (2)  of  §  8  and  (1)  of  §  9 
as  the  solutions  of  their  I'espective  kinds  of  problems,  and  to  follow  the 
same  methods  in  solving  similar  numerical  problems,  rather  than  to 
memorize  them  and  substitute  numerical  values  for  ri,  x^,  etc. 


(1) 


10.     Relation  between  Cartesian  and  Polar  Coordinates. 

As  the  position  of  the  same  point  may  be  determined 
either  by  x  and  y  or  by  r  and  6,  there  must  be  some  equations, 
connecting  x  and  y  with  r  and  6. 

Let  the  origin  of  polar  coordinates  be  the  origin  of  rect- 
angular coordinates,  and  the  initial  line  of  polar  coordinates 
coincide  with  OX. 

Y 


Fig.  13. 


THE    POINT.  17 


If  the  rectangular  coordinates  of  P  (r,  0)  are  x  and  y,  by 
Trigonometry, 

X  =  r  cos  0, 
y  =  r  sin  9- 

From  [7],  we  may  derive  the  formulas 


[7J 


r  =  Vx-  +  y-, 

e-tan-'^-  ^^^ 

X 

Ex.  1.     Find  the  rectangular  coordinates  of  P(o,  30°) ;  by  [7], 

3V3 
X  =  3  cos  30°  =  — — , 

y  =  3  sin  30°  =  -  ■ 

Ex.  2.     Find  the  polar  coordinates  of  P(—  2,  +  3) ;  by  [8], 
r  =1  V4  +  9  =  Vl3, 
e  =  tan-i  (—  3)  =  180°  -  tan-i  (l) 

=  180° -50°  19'"=  123°  41'. 

If  the  angle  XOY  between  the  axes  of  the  Cartesian  coordi- 
nates is  w,  it  is  not  difficult  to  prove  the  following  formulas  : 

X  =  r  cos  6  —  r  sin  0  ctn  w,       ^ 

r  sin  ^  y  (1) 


y  = 


sm  w 


r  =  Vx"  +  y'  +  ^  xy  cos  to,       ^ 
|_x  H-y  cos  wj         J 


(2) 


EXAMPLES. 


1.  Find  the  length  and  the  slope  of  the  line  joining  (3,  4) 
and  ( —  5,  —  6). 

2.  Find  the  length  and  the  slope  of  the  line  joining  ( —  5,  2) 
and  (3,  —  1). 

3.  Find  the  lengths  and  the  slopes  of  the  sides  of  the  triangle 
of  which  the  vertices  are  (1,  2),  (—3,  4),  and  (2,-3). 


18  PLANE    ANALYTIC    GEOMETRY. 

4.  Find  tlie  lengths  and  the  slopes  of  the  sides  of  the  tri- 
angle of  which  the  vertices  are  ( —  3,  —  4),  (0,  — 2),  and  (4,  3), 

5.  Prove  that  the  point  (1,  5)  is  on  the  line  joining  the 
points  (0,  2)  and  (2,  8)  and  is  equally  distant  from  them. 

6.  Prove  the  triangle  having  its  vertices  at  the  points  (2, 
2),  (—  2,  —  2),  and  (2  \/3,  —  2  V3)  an  equilateral  triangle. 

7.  Find  the  perimeter  of  the  triangle  having  as  its  vertices 
the  points  (a,  b),  ( — a,  b),  and  ( — a,  —  b). 

8.  Prove  the  quadrilateral  having  its  vertices  at  the  points 
(2,  3),  (—1,  2),  (—2,-3),  and  (1,-2)  a  parallelogram. 

9.  Show  that  the  points  (2,  3),  (—2,  3),  (—2,  —3),  and 
(2,  —  3)  lie  upon  a  circle  of  which  the  centre  is  the  origin. 

10.  Find  the  coordinates  of  P  such  that  PiP  :  PP2  =  3  :  5, 
where  Pj  is  (2,  2)  and  P^  is  (4,  3). 

11.  Find  the  coordinates  of  the  point  which  divides  the  dis- 
tance from  ( — 3,  4)  to  (—  5,  — 6)  in  the  ratio  5  :  7. 

12.  Find  the  coordinates  of  the  point  f  of  the  distance  from 
(3, -4)  to  (-2,  5). 

13.  Find  the  coordinates  of  the  points  trisecting  the  line 
joining  (2,  3)  and  (4,  5). 

14.  Find  a  point  which  divides  the  distance  from  the  origin 
to  the  point  (4,  3)  externally  in  the  ratio  6  :  5. 

15.  On  the  straight  line  passing  through  (1,3)  and  (4,  —  6) 
find  a  point  such  that  its  distances  from  the  given  points 
respectively  shall  be  in  the  ratio  —  |. 

16.  To  what  point  must  the  line  drawn  from  (—  2,  —  3) 
to  (3,  7)  be  extended  in  the  same  direction  that  its  length  may 
be  trebled  ? 

17.  Given  the  three  points  A  (4,-1),  B  (1,-4),  and  C 
(—  1,  —6)  upon  a  straight  line.  Find  a  fourth  point  D  such 
fi    .  AD  AB 

'^^"*  DC  =^  -  BC- 

18.  One  extremity  of  a  line  is  at  tlie  point  (3,  —5)  and  a 
point  I  of  the  distance  to  the  other  extremity  is  (4,  3).  Find 
the  other  extremity  of  the  line. 


THE    POINT.  19 

19.  Find  the  lengths  of  the  medial  lines  of  the  triangle 
(-5,3),  (-3,  7),  (1,-1). 

20.  How  far  is  the  point  bisecting  the  line  joining  the 
points  (5,  5)  and  (—  3,  7)  from  the  origin  ?  What  is  the 
slope  of  this  last  line  ? 

21.  Find  the  length  and  the  slope  of  the  line  joining  the 
points  ( — 1,  3)  and  (2,  5),  tlie  axes  being  oblique  and  to  =  60°. 

22.  Find  the  area  of  the  triangle  (0,  0),  (3,  1),  (—2,  0). 

23.  Find  the  area  of  the  quadrilateral  (—2,  3),  (—3,  —4), 
(5,-1),  (2,2). 

24.  Find  the  area  of  the  pentagon  (1,  2),  (3,-1),  (6,-2), 
(2,  5),  (4,  4). 

25.  What  is  the  ratio  of  the  areas  of  the  triangles  (2,  3), 
(1,  2),  (3,-1)  and  (5,-4),  (2,-1),  (-1,  3)? 

26.  The  vertices  of  a  triangle  are  the  points  (1,  2),  (3,  —  5) 
and  ( — 2,  1).  What  is  its  area?  What  is  the  area  of  the 
triangle  formed  by  joining  the  middle  points  of  its  sides  in 
succession  ? 

27.  If  the  angle  between  the  axes  is  30°,  plot  the  points 
(1,  2),  ( — 2, — 4),  and  (3,  —  5),  and  find  the  perimeter  of  the 
triangle  formed. 

28.  Find    the    rectangular    coordinates    of    the    following 

points:   (2,  30°);   (-3,45°);   (l-"!)- 

29.  Find  the  polar  coordinates  of  the  following  points  : 
(2,5);  (3,4);  (-5,12). 

30.  Find  the  lengths  of  the  sides  and  the  area  of  the 
triangle  (5,  15°),  (3,  75°),  (6,  135°). 

31.  Find  the  perimeter  and  the  area  of  the  triangle  (1,  30°), 
(3,  60°),  and  (5,  90°). 

32.  Show  that  the  distance  between  the  two  points  (r,,  d{) 
and  (ro,  O.2)  is  a  maximum  when  6..  —  ^i  =  180°,  and  a  mini- 
mum when  Oi=^0.2,  ri  and  r.,  remaining  unchanged. 

33.  In  the  triangle  ABC,  A  (1,  2),  B  (3,  -2),  C  (2,  5),  a 
line  is  drawn  bisecting  the  adjacent  sides  AB  and  BC.  Prove 
this  new  line  to  be  parallel  to  AC  and  half  as  long. 


20  PLANE    ANALYTIC    GEOMETRY. 

34.  Prove  that  the  area  of  a  triangle,  in  oblique  coordinates, 
is  K^iYa  — ><2yi+x2y3  — X3y2  +  X3yi  — xiYa)  sin  w,  0)  being  the 
angle  between  the  axes. 

35.  Pind  the  coordinates  of  a  point  equally  distant  from 
the  points  (2,  3),  (—2,  —3),  and  (1,  4). 

36.  A  point  is  distant  7  units  from  the  origin  and  equally 
distant  from  the  two  points  (1,  2)  and  ( —  2,  — 1).  What  are 
its  coordinates? 

37.  A  point  is  distant  5  units  from  the  origin,  and  the 
slope  of  the  line  joining  it  to  the  origin  is  f .  What  are  its 
coordinates  ? 

38.  Assuming  that  the  medial  lines  of  a  triangle  meet  in  a 
point  I  of  the  distance  from  each  vertex  to  the  opposite  side, 
find  this  medial  point  of  the  triangle  (3,  8),  ( — 2,  7),  (1,  — 4). 

39.  Show  that  the  medial  point  of  the  triangle 

(x„  y,).  (X.  y.),  (X.,  ya)  is  (''■+;'+^  y+i'+y-y 

40.  Show  that  the  lines  from  the  vertices  to  the  medial 
point  of  the  triangle  (3,  — 8),  (—4,  6),  (7,  0)  divide  it  into 
three  triangles  of  equal  area. 

41.  Show  that  the  line  joining  any  two  vertices  of  the  tri- 
angle (xx,  yi),  (x2,  y2),  (xs,  ya)  to  the  medial  point  form  with 
the  adjacent  side  of  the  triangle  another  triangle  of  one  third 
the  area  of  the  given  triangle. 

42.  Given  four  points  Pi  (xi,  yi),  P^ix^,  ya),  Psi^s,  ya), 
P4(x4)y4);  fiiid  the  point  half  way  between  Pi  and  Pj,  then 
the  point  one  third  of  the  distance  from  this  point  to  Pg,  and 
finally  the  point  one  fourth  of  the  distance  from  this  point  to 
P4.  Show  that  the  order  in  which  the  points  are  taken  does 
not  affect  the  result. 

43.  Prove  analytically  that  the  line  joining  the  middle 
points  of  two  sides  of  a  triangle  is  parallel  to  the  third  side 
and  equal  to  one  half  of  it. 

44.  Prove  analytically  that  a  line  which  divides  two  sides 
of  a  triangle  proportionally  is  parallel  to  the  third  side. 


THE    POINT.  21 

45.  Prove  analytically  that,  if  in  any  triangle  a  median  be 
drawn  from  the  vertex  to  the  base,  the  sum  of  the  squares  of 
the  other  two  sides  is  equal  to  twice  the  square  of  half  the 
base,  plus  twice  the  square  of  the  median. 

46.  Prove  analytically  that  the  diagonals  of  a  parallelogram 
bisect  each  other. 

47.  Prove  analytically  that  the  lines  joining  the  middle 
points  of  the  adjacent  sides  of  any  quadrilateral  form  a  paral- 
lelogram. 

48.  Prove  analytically  that  the  above  parallelogram  (Ex. 
47)  is  equivalent  to  one  half  the  quadrilateral. 

49.  Prove  analytically  that  in  any  right  triangle  the  straight 
line  drawn  from  tlie  vertex  to  the  middle  point  of  the  hypot- 
enuse is  equal  to  one  half  the  hypotenuse. 

50.  Show  that  the  sum  of  the  squares  on  the  four  sides  of 
any  quadrilateral  is  equivalent  to  the  sum  of  the  squares  on 
the  diagonals,  together  with  four  times  the  square  of  the  line 
joining  the  middle  points  of  the  diagonals. 


CHAPTER    n. 

LOCI. 

11.     Relation  between  Equation  and  Locus. 

We  have  seen,  in  the  previous  chapter,  that  two  conditions 
are  necessary  to  fix  the  position  of  a  point  in  a  plane  ;  for 
example,  x  ^  a,  y  =  b.  If  only  one  of  these  conditions  is 
given,  the  point  is  not  fully  determined,  but  may  lie  any- 
where upon  a  certain  straight  line  or  curve.  For  example, 
X  =  a  gives  a  condition  which  is  true  for  all  points  upon  a 
straight  line  parallel  to  the  axis  of  y  and  at  a  distance  of  a 
units  from  it.  We  say  that  the  equation  x  =  a  represents  this 
straight  line.  In  the  same  way,  the  equation  y  =  b  represents 
a  straight  line  parallel  to  the  axis  of  x,  at  a  distance  of  b 
units.  As  particular  cases,  x  ==  0  represents  the  axis  of  y  and 
y  =  0  represents  the  axis  of  x. 

The  above  are  simple  examples  of  the  interpretation  of 
an  algebraic  equation  in  Analytic  Geometry.  In  the  same 
manner,  any  equation  between  x  and  y  will  be  true  for  the 
coordinates  of  an  infinite  number  of  points,  which  will  be 
found  to  lie  upon  a  certain  curve  or  locus.  We  say  that  the 
equation  represents  the  locus,  or  that  it  is  the  equation  of  the 
locus.  Thus,  the  equation  x  =  y  is  true  for  the  coordinates  of 
all  points  which  lie  upon  the  straight  line  bisecting  the  angles 
of  the  first  and  the  third  quadrants  between  the  coordinate 
axes.  It  is,  therefore,  the  equation  of  this  straight  line. 
Thus,  also,  x^  +  y-  ^  25  is  the  equation  of  a  circle,  the  centre 
of  which  is  at  the  origin  and  the  radius  of  which  is  5  ;  for  the 
equation  is  evidently  true  for  the  coordinates  of  all  points 
upon  this  circle,  and  for  those  of  no  other  points.  This  rela- 
tion between  the  equation  and  the  locvis  is  fundamental  for 


LOCI. 


23 


the  work  of  Analytic  Geometry,  and  because  of  its  great  im- 
portance we  restate  it  for  easy  reference  as  follows  : 

All  2^oints  the  coordinates  of  ivhich  satisfy  a  given  equation 
lie  upon  a  certain  curve,  called  the  locus  of  the  equation  ;  and, 
conversely,  if  a  point  lies  upon  the  locus,  its  coordinates  satisfy 
the  equation  of  the  locus. 

12.     Equation  of  the  First  Degree. 

It  folloAvs  from  the  the  above  that  the  locus  of  an  equation 
may  be  empirically  constructed.  We  need  only,  by  trial,  to 
find  a  sufficient  number  of  points  the  coordinates  of  which 
satisfy  the  equation,  in  order  to  obtain  a  dotted  outline  of  the 
curve.  The  rest  of  the  curve  may  then  be  tilled  in  free-hand, 
provided  the  points  are  near  enough  together. 

Ex.     Let  us  take,  for  example,  the  equation 


24  PLANE   ANALYTIC    GEOMETRY. 

We  may  find  any  number  of  points  on  the  locus  by  giving  to  x  succes- 
sively any  number  of  arbitrary  values  and  computing  the  corresponding 
values  of  y  from  the  equation. 

For  example,  giving  x  the  value  —  3,  we  have  —  6  +  oy  =  0,  whence 
y  =  -1.  The  values,  x  =  —  3,  y  =  4,  satisfy  the  equation,  and  hence  the 
point  (—3,  4)  lies  upon  the  locus.  In  this  way  we  find  the  following 
points:  (-3,  4),  (-2,  -\%  (-  1,  »),  (0,  2),  (1,  f),  (2,  |),  (3,  0),  (4,  -  |). 

If  these  points  are  plotted,  it  will  be  found  that  a  straight  line  can  be 
drawn  through  them,  as  in  Fig.  14. 

We  assume,  in  the  example,  that  the  locus  is  a  straight  line. 
The  student  should  motice,  however,  that  this  is  really  an 
assumption.  Our  work  gives  us  no  information  concerning 
any  points  of  the  locus  except  those  actually  found.  We  will 
draw  a  smooth  curve  through  these  points,  however,  thus 
obtaining  an  approximation  to  the  shape  of  the  curve,  the 
accuracy  of  the  approximation  depending,  in  general,  upon 
the  nearness  together  of  the  points  actually  found. 

In  the  next  chapter,  we  shall  prove  that  any  equation  of 
the  first  degree  represents  a  straight  liyie.  Hence,  if  we  wish  to 
find  the  locus  of  an  equation  of  the  first  degree,  it  is  necessary 
to  find  only  two  points  and  to  draw  a  straight  line  through 
them.  The  two  points  most  readily  found  are  those  in 
which  the  straight  line  cuts  the  two  axes.  The  point  B  in 
which  the  line  cuts  the  y-axis  is  the  point  for  which  x  =  0  ; 
i.e.,  in  the  example,  the  point  (0,  2).  Similarly,  the  point  A 
in  which  the  line  cuts  the  x-axis  is  the  point  for  which  y  ^  0  ; 
i.e.,  the  point  (3,  0).  The  distances  OA  and  OB  are  called 
the  intercepts  on  the  axes  of  x  and  y  respectively. 

13.     Equations  of  Degree  Higher  than  the  First. 

If  the  equation  to  be  considered  is  of  a  degree  higher  than 
the  first,  the  work  becomes  more  difficult.  It  is  not  easy  to 
give  specific  directions  which  will  apply  to  all  cases,  but  the 
following  plan  of  work  may  be  followed  in  the  simpler  cases 
at  least. 


LOCI.  25 

(1)  Find  the  points  in  which  the  curve  cuts  the  coordinate 
axes. 

(2)  Solve  the  equation  for  one  of  the  coordinates  in  terms 
of  the  otlier,  let  us  say  for  y  in  terms  of  x. 

(3)  Find  what  values  of  x,  if  substituted  in  the  equation, 
make  y  imaginary.  These  values  do  not  correspond  to  any 
real  points  of  the  curve. 

(4)  Assume  values  of  x,  not  excluded  by  (3),  and  compute 
corresponding  values  of  y. 

(5)  Plot  the  points  thus  found  and  draw  a  smootli  curve 
through  them. 

The  above  directions  are  given  as  hints  to  the  student  and 
not  as  rigid  rules  to  be  invariably  followed.  Special  prob- 
lems often  require  modification  of  the  method.  It  should  be 
remembered  that  the  whole  object  is  to  find,  by  trial,  enough 
points  of  the  curve  to  outline  it,  and  the  method  whicli  will 
do  this  most  quickly  is  the  best.  In  particular,  we  should 
notice  tliat  it  will  sometimes  be  found  more  convenient  to 
solve  the  equation  for  x  instead  of  y.  In  that  case,  tlie  letters 
X  and  y  are  to  be  interchanged  in  reading  directions  (2),  (3), 
and  (4).  No  definite  instruction  can  be  given  as  to  how  near 
together  the  points  should  be  taken.  When  the  curve  changes 
its  direction  slowly,  they  may  be  far  apart ;  but  if  the  shape 
of  the  curve  is  doubtful  or  peculiar  at  any  place,  the  points 
must  be  taken  close  enough  to  make  the  proper  shape  per- 
fectly clear. 

Ex.   1.   Take,  as  the  first  example, 

4x-+  9y-=36. 

(1)  riacing  X  =  0,  we  have 

0y2  =  m, 
y   =±2. 
Hence  the  curve  cuts  the  axis  of  y  in  the  two  points  (0,  2)  and  (0,  —  2). 

Placing  y  =  0,  we  find  x  =  ±  .'].  Hence  the  curve  cuts  the  axis  of  x 
in  the  two  points  (.'1,  0)  and  (—  3,  0). 

(2)  The  equation  is  readily  put  into  the  form 


y  =  ±  ^  V9  -  x2  =  ±  2  V(3  -  X)  (3  +  X). 


26 


PLANE   ANALYTIC   GEOMETRY. 


(3)  It  is  apparent  that,  it  x  <  —  o  or  x  >  ;J,  tlie  quantity  under  the 
radical  sign  is  negative,  and  hence  y  is  imaginary.  Therefore  there  can 
be  no  point  of  the  curve  above  or  below  that  portion  of  the  x-axis  which 
lies  to  the  left  of  —  3  or  to  the  right  of  +  3. 

(4)  Using  values  of  x  which  lie  between  —  3  aud_+  3,  we  find  the 
following  points  of  the  curvej  (-3,  0),  (-2,  ±|V5),  (-1,  ±|V8), 
(0,  ±2),  (1,  ±|V8),  (2,  ±  3V5),  (3,  0). 

(5)  The  plotting  of  the  points  leaves  the  shape  of  the  curve  somewhat 
doubtful  in  the  extreme  right  and  left  hand  portions.  Therefore  we  com- 
pute also  the  points  (—2^,  ±  Wll),  (2.1,  ±iVll),  and  then  draw  the 
curve  through  all  the  points  found.     The  curve  is  an  ellipse.    (Fig.  15.) 


We  have  seen  that  the  curve  cannot  lie  to  the  left  of  x  =  —  3  or  to 
the  right  of  x  =  +  3.  Similarly,  we  may  show  that  it  cannot  lie  above 
y  =  +2  or  below  y  =  —  2.  This  we  do  by  solving  the  equation  for  x 
We  have,  then, 

X  =  ±  I V4  -  y2  =  ±  3  V(2  -  y)  (2  +  y). 

From  this  it  appears  that  x  is  imaginary,  if  y  is  >»2  or  <  —  2.  This 
finding  of  the  limits  of  the  curve  in  the  vertical  as  well  as  the  horizontal 
direction  is  not  necessary  to  the  plotting,  but  is  often  useful. 


LOCI.  27 

It  will  be  noticed  that  the  curve  is  symmetrical  with  respect  to  the 
axis  of  X. 

This  follows  from  the  fact  that  to  each  value  of  x  correspond  two 
values  of  y,  equal  in  absolute  magnitude  but  opposite  in  sign.  Similarly, 
since  to  any  value  of  y  correspond  two  values  of  x,  equal  in  magnitude 
but  opposite  in  sign,  the  curve  is  symmetrical  with  respect  to  the  axis  of 
y.     It  is  useful  in  plotting  to  notice  such  symmetry  when  possible. 

Ex.  2.     Take,  as  a  second  example,  the  equation 
4x2  _  oy'2  _  i6x  +  18y  -  29  =  0. 

(1)  When  x  =  0, 

9y2  —  18y  =  —  29, 

3  ±  V-'20 
whence  y  — ^ . 

This  is  imaginary.     Hence  the  curve  does  not  cut  the  axis  of  y. 
When  y  =  0, 

4x2  _  lox  =  29, 

4±3V5       .  ,,  ,  „, 

or  X  = =  5.35  or  —  1.35. 

Hence  the  curve  cuts  the  axis  of  x  in  the  points  (5.35,  0)  and 
{-  1.35,  0). 

(2)  By  solving  the  equation  for  y,  we  obtain 


3±2Vx2-4x-5      3±2V(x-5)  (x  +  1)     . 

y= 3 = 3         • 

(3)  The  value  of  y  will  be  imaginary  when  the  product  (x  —  5)  (x  +  1) 
is  negative;  i.e.,  when  x  has  values  which  cause  the  factors  to  have  oppo- 
site signs.  Such  values  of  x  are  seen  by  inspection  to  be  those  which  are 
>  —  1  and  <  +  5.  Hence  there  is  no  point  of  the  curve  above  or  below 
the  portion  of  the  x-axis  between  —  1  and  +  5. 

(4)  Giving  x  in  succession  the  values  —4,  —3,  —2,  —  1,  5,  0,  7,  8, 
we  find  the  following  points  of  the  curve  : 

(-4,:±W.).   (-,s,lf^),    (-.■^'),    (-,.). 

(5)  The  plot  results  as  in  Fig.  10. 

There  are  no  limits  to  the  value  of  y.     For,  if  we  solve  for  x,  we  have 

4±3Vy2-2y  +  5 
x  = 

and  no  value  of  y  will  make  the  quantity  under  the  radical  sign  negative. 


28 


PLANE    ANALYTIC    GEOMETRY. 


As  regards  symmetry,  the  curve  is  evidently  not  symmetrical  with 
respect  to  the  coordinate  axes.     But  from  the  equation 


y  =  l±|V(x-5)  (x+l), 

it  appears  that  the  curve  is  symmetrical  with  respect  to  the  line  y  =  1. 
Similarly,  from  

X  =  2  ±  I  Vy2  —  2y  +  5, 

it  appears  that  the  curve  is  symmetrical  with  respect  to  the  line  x  =  2. 


^ 

/ 

5 

N 

s 

4 

/ 

/' 

s 

\ 

3 

/ 

/ 

^ 

^ 

2 

/ 

^ 

\ 

1 

/ 

y' 

-5 

-4 

-3 

-2/ 

■10 

1 

2 

3 

4 

5 

\e 

7 

8 

/ 

/ 

-1 

\ 

^ 

y 

/ 

-2 

\ 

s 

? 

/ 

-3 

s 

^ 

-4 

-5 

Fig.  16. 


14.     Transcendental  Equations. 

The  equation  of  a  locus  is  not  necessarily  algebraic,  but 
may  be  transcendental ;  that  is,  it  may  involve  the  non- 
algebraic  functions.  A  few  examples  will  make  clear  the 
methods  of  handling  such  equations. 


Ex.  1.     Take  the  equation 


y  =  sni  X. 


We  consider  the  angle  x  expressed  in  circular  measure,  and  lay  off 
first  the  distance,  7r  =  3.1416,  and  its  multiples.     Each  division  of  the 


LOCI. 


29 


axis  we  then  subdivide  into  convenient  intervals,  say  sixtlis,  and  obtain 
thus  upon  tlie  x-axis  the  points 

It     Tt     7t     2Tt     bit  ItC     ^TC     ?>Tt       ^ 

—,  —1  — '  — ^'  — '  7t,  —  '  -^'  — ^'  etc. 
G'  3    2     3      G       '    G       3      2 

By  means  of  a  table  of  natural  sines,  the  value  of  y  corresi>onding  to 
each  X  is  readily  found.     The  plot  results  as  in  Fig.  17  : 


It  is  necessary  to  plot  only  that  portion  of  the  curve  between  x  =  0 
and  X  =  TT.  For  sin  (x  ±  tt)  =  —  sin  x,  and  sin  (x  ±  2;r)  =  sin  x.  Hence 
the  curve  y  =  sin  x  is  composed  of  an  indefinite  number  of  arches  extend- 
ing both  to  the  left  and  to  the  right,  the  arches  being  alternately  above 
and  below  the  axis  of  x,  and  each  arch  being  of  the  same  shape  as  that 
between  x  =  0  and  x=  tt. 

Ex.  2.     Consider,  as  another  example, 

y  =  5  log  X. 

By  use  of  a  table  of  logarithms,  any  number  of  points  are  readily 
found  with  the  following  result,  as  shown  in  Fig.  18. 


30 


PLANE    ANALYTIC    GEOMETRY. 


Fig.  18. 

It  is  to  be  noticed  that  the  curve  approaches  the  axis  of  y  in  the  nega- 
tive direction,  but  never  reaches  it.  Tliis  arises  from  the  fact  that  the 
logarithm  of  a  quantity  increases  negatively  without  limit  as  the  quantity 
approaches  zero. 

It  is  also  to  be  noticed  that  we  have  no  portion  of  the  curve  corre- 
sponding to  the  negative  values  of  x,  owing  to  the  fact  that  the  logarithm 
of  a  negative  quantity  is  not  a  real  quantity. 


15.     Equations  in  Polar  Coordinates. 

The  equation  of  a  locus  may  also  be  given  in  polar  coordi- 
nates (r,  6).  To  plot  sucli  a  curve,  it  will  be  found  convenient 
to  lay  off  first  a  number  of  radii  at  convenient  angles.  Then 
the  corresponding  values  of  r  may  be  computed  from  the 
equation,  perhaps  with  aid  of  a  table  of  trigonometric  func- 
tions. These  values  of  r  are  then  laid  off  upon  the  corre- 
sponding radii,  with  careful  attention  to  the  meaning  of  the 
negative  sign,  as  explained  in  §  7. 

For  example, 

r  —  2a  sin  6. 

We  compute  the  following  table  of  corresponding  values  : 


LOCI. 


31 


61  =  0°,  r  =  0;  6=  15°,  r=  .52a;  61  =  30°,  r  =  a;  ^  =  45°,  r  =  1.41a 
61  =  60°,  r=  1.73a;  61  =75°,  r=  1.93a;  61  =  90°,  r  =  2a;  6i  =  105°,  r  =  1.93a 
61  =  120°,  r  =  1.73a;  6*=  135°,  r=  1.41a;  tf  =  150°,  r=a;  6i  =  165° 
r  =  .52a;  ^  =  180°,  r  =  0;  6*  =  195°,  r=— .52a;  ^=210°,  r=— a 
6*  =  225°,  r=  — 1.41a;  6i  =  240°,  r=- 1.73a;  61  =  2.55°,  r=  — 1.93a 
(?=270°,  r  =  -2a;  6*  =  285°,  r=  — 1.93a;  61=300°,  r=  — 1.73a 
^  =  315°,  r=- 1.41a;  61=330°,  r=  — a;  d*  =  345°,  r=-.52a 
e  -  300°,  r  =  0. 

The  plot  results  as  in  Fig.  19. 


For  angles  in  the  third  and  the  fourth  quadrants  r  is  negative,  and  is, 
therefore,  measured  along  the  backward  extension  of  the  terminal  line. 
For  example,  if  ^=195°,  r=  —.52a.  We  measure  .52a  units  away 
from  the  195°  point  and  toward  the  15°  point.  We  obtain,  therefore, 
the  same  point  of  the  curve  for  9  =  195°  as  we  obtained  for  6=  15°.  In 
the  same  way,  all  angles  of  the  third  and  the  fourth  quadrants  give  the 
same  points  as  angles  of  the  first  and  the  second  quadrants.  Hence  we 
obtain  the  curve  above  the  initial  line  twice.     The  curve  is  a  circle. 


32  PLANE    ANALYTIC    GEOMETRY. 

16.     Locus  Defined  by  Geometric  Property. 

A  locus  is  often  defined  by  means  of  a  property  which  is 
common  to  all  points  upon  the  locus,  but  to  no  other  point. 
The  object  of  Analytic  Geometry  is  then  to  express  this  prop- 
erty by  means  of  an  algebraic  equation  between  the  coordi- 
nates of  any  point  upon  the  locus,  and  then  to  find  the  shape 
and  the  properties  of  tlie  locus  from  the  equation. 

For  example,  let  it  be  required  to  find  the  locus  of  a  point  at  a  dis- 
tance of  5  units  from  the  origin.  By  [1],  §  3,  the  distance  of  a  point 
(x,  y)  from  the  origin  is   Vx^  +  y^.     For  the  locus,  therefore,  we  have 


Vx-  +  y-  =  5, 
or  x2  -f-  y2  =  25. 

This  is  the  equation  of  a  circle  of  which  the  centre  is  at  the  origin  and 
the  radius  is  5. 

Again,  let  it  be  required  to  find  the  locus  of  a  point,  the  sum  of  the 
distances  of  which  from  the  points  (—  1,  0)  and  (1,  0)  is  equal  to  6.  By 
[1],  §  3,  we  have 

V(x-h  1)2 -hy^  +  V(x  — l)2-}-y2  =  6, 
which  reduces  readily  to 

8x2  ^  f,y2  =:  72. 

This  locus  is  readily  plotted  and  found  to  be  similar  to  that  of  Ex.  1, 
§  13. 

17.     Intersection  of  Loci. 

It  is  often  necessary,  when  the  equations  of  the  two  loci 
are  given,  to  find  the  coordinates  of  the  point  or  points  in 
which  the  loci  intersect.  Now,  a  point  of  intersection  of  two 
loci  is  a  point  which  lies  upon  both,  and  hence  by  §  11  its 
coordinates  satisfy  the  equation  of  each.  The  problem  is, 
therefore,  to  find  those  values  of  x  and  y  which  satisfy  both 
equations,  or  in  other  words  : 

To  find  the  coordinates  of  the  jjoints  of  intersection  of  two  loci, 
solve  their  equations  for  simultaneous  values  of  x  and  y. 

Two  cases  are  of  special  importance  in  elementary  work. 

Case  I.      When  both  equations  are  of  the  first  degree. 


LOCI. 


33 


There  is,  in  general,  one  and  only  one  pair  of  values  which 
satisfies  the  equations.  The  two  loci  intersect,  therefore,  in 
one  point,  as  is  evident  if  we  accept  the  statement  in  §  12 
that  each  locus  represents  a  straight  line. 

For  example,  consider  the  equations 

2x  +  .3y  =  5, 
4x  —  5y  =  8, 

the  solution  of  which  is  x  =  f|,  y  =  ^\.     The  loci  intersect  in  the  point 
(II)  T'r)'  ^s  shown  in  Fig.  20. 


Fig.  20. 

An  exceptional  case  arises  when  the  two  equations  are 
contradictory  ;  for  example, 

2x  +  %  =  4, 
2x  +  ;5y  =  7. 

Since  we  can  find  no  values  of  x  and  y  common  to  these  two 
equations,  it  follows  that  the  two  loci  do  not  intersect.  In 
fact,  a  plot  shows  that  the  equations  represent  parallel 
straight  lines. 


34  PLANE    ANALYTIC    GEOMETRY. 

A  rigorous  discussion  of  the  principle  here  involved  may  be  given  as 
follows : 

Let  ax  +  by  =  c,  (1) 

and  a'x  +  b'y  =  c',  (2) 

be  two  equations,  each  representing  a  straight  line.     By  elimination,  we 
find  (ab' —  a'b)x  =  b'c  —  be', 

and  (ab' —  a'b)y  =  ac'  —  a'c. 

Hence,  unless  ab'  —  a'b  =  0,  we  find  as  the  solution  of  the  two  equa- 
tions, 

_  b'c  —  be' 

ab' — a'b'  /^\ 

_  ac'  —  a'c 
^  ~  ab'-a'b ' 

and  these  are  the  coordinates  of  the  point  of  intersection  of  the  two  lines. 

If  ab'  —  a'b  =  0,  the  above  solution  is  impossible.  A  meaning  may 
be  given  to  equations  (3),  however,  by  use  of  the  method  of  limits.  To 
do  this,  let  us  suppose,  first,  that  the  coefficients  of  (1)  and  (2)  are  such 
that  ab'—  a'b  is  a  very  small  quantity,  although  not  zero.  The  values 
of  X  and  y  in  (3)  are  then  possible,  but  are,  in  general,  very  large.  If, 
now,  the  expression  ab'  —  a'b  can  be  made  to  approach  zero  as  a  limit, 
the  values  of  x  and  y  will  increase  without  limit,  except  in  the  case  men- 
tioned below.  Now  ab'  —  a'b  can  be  made  to  approach  zero  in  various 
ways;  for  example,  we  may  keep  (1)  unchanged,  and  alter  step  by  step 
the  coefficients  of  (2).  If  each  step  is  accompanied  by  a  plot,  we  see  that 
the  line  (1)  is  kept  fixed,  but  the  position  of  (2)  is  altered.  The  point  of 
intersection  of  the  two  lines  is  at  each  step  more  remote.  Hence  the  case 
ab'  —  a'b  =  0  may  be  regarded  as  the  limiting  case  of  two  lines,  the  point 
of  intersection  of  which  has  receded  indefinitely  from  the  origin.  But 
the  limiting  position  of  two  such  lines  is  evidently  one  of  parallelism,  and 
hence  it  follows  that  if  ab'  —  a'b  =  0,  the  two  lines  are  parallel.  (See 
also  §  35.) 

The  above  reasoning  fails,  if  at  the  same  time  that  ab'  —  a'b  =  0,  the 
numerators  of  the  fractions  in  (3)  are  also  zero.  But,  in  that  case,  we 
have  evidently 

a  _  b  _  c 
a'       b'      c' ' 

and  the  equations  (1)  and  (2)  are  identical.     We  have,  therefore,  only 
one  line  in  the  problem. 


LOCI. 


35 


18.     Intersection  of  Loci. 

Case  II.  When  one  equation  is  of  the  first  degree  and  the 
other  of  the  second. 

We  shall  find  it  convenient  to  eliminate  by  substituting  in 
the  quadratic  equation  the  value  of  one  of  the  unknown  quan- 
tities as  given  in  terms  of  the  other  by  the  simple  equation. 

Let  us  suppose  we  eliminate  y  in  this  way.  There  results, 
in  general,  a  quadratic  equation  in  x,  let  us  say 

ax'-+bx  +  c  =  0,  (1) 

the  roots  of  which  are  the  abscissas  of  the  required  points  of 
intersection.  When  the  values  of  x  have  been  found,  the 
corresponding  values  of  y  are  found  hy  siihsthuting  in  the 
simple  equation.  In  the  solution  of  the  above  quadratic 
equation,  one  of  three  cases  will  arise,  as  illustrated  in  the 
following  examples  : 

Ex.  1. 

(1) 


To  find  the  intersections  of 

y  -t-  2x  =  3, 


and 


X-  +  y2  =  5. 


(2) 


36 


PLANE    ANALYTIC    GEOMETRY. 


Eliminating  y  by  substitution  from  (1)  into  (2),  there  results 
5x2-12x  +  4  =  0, 
which  gives  x  =  2  or  |.  The  corresponding  y's  found  from  (1)  are  —1  or  y-. 
Therefore  the  points  of  intersection  are  (2,  —  1)  and  (|,  -U-).     (Fig.  21.) 
Ex.  2.     To  find  the  intersections  of 

y  +  2x  =  5         and         x''^  +  y-  =  5. 
Y 
Je 


Eliminate  y,  and  there  results 

5x2  _  20x  +  20  =  0. 
This  equation  has  the  equal  roots  2  and  2,  to  which  correspond  the 
equal  y's,  1  and  1.     The  loci  are  tangent  at  the  point  (2,  1).     (Fig.  22.) 

Ex.  3.     Find  the  intersections  of 

y  +  2x  =  7         and         x^  +  y2  =  5. 

Eliminating  y,  we  have 

5x2  _  28x  +  44-0, 

14±2V-6 

whence  x  = • 

5 

Since  this  is  imaginary,  the  two  loci  do  not  intersect.     (Fig.  23.) 


LOCI. 


37 


We  may  now  state  the  three  cases  which  may  occur  in  the 
solution  of  the  general  equation  (1)  on  p.  35,  as  follows : 

(1)  If  b^  —  4ac  >  0,  the  two  roots  are  real  and  uneqnal. 
The  two  loci  intersect  in  tivo  distinct  points. 

(2)  If  b"^  —  4ac:=0,  the  ttvo  roots  ore  real  and  equal.  The 
tivo  points  of  intersection  become  coincident ;  the  two  loci  are 
therefore  tangent. 


38 


PLANE   ANALYTIC   GEOMETRY. 


(3)  If  h"^  —  4ac  ■<  0,  the  roots  are  imaginary.  The  two  loci 
fail  to  intersect. 

It  is  sometimes  said  in  such  a  case  as  the  last,  that  tlie  loci 
intersect  in  "imaginary  points."  The  beginner  will  under- 
stand this  as  equivalent  to  saying  that  the  loci  do  not  inter- 
sect at  all.  The  meaning  of  the  phrase  is  that  imaginary 
values  of  x  and  y  may  be  found  which  satisfy  both  equations, 
and  the  advanced  student  finds  advantage  in  recognizing  this 
fact. 


19.     Limiting  Cases  of  Intersection. 

In  the  process  of  eliminating  y  as  outlined  in  the  last 
article,  it  may  happen,  in  exceptional  cases,  that  the  term 
containing  x^  cancels  out.  The  loci  then  intersect  in  only 
one  point. 

Ex.  1.     Consider 


By  elimination, 
whence 


24x  -  72  =  0, 
X  =  3,     y  =  0. 


The  loci  intersect  in  one  point  (3,  0).     (Fig.  24.) 


LOCI. 


39 


Again,  it  may  happen  that  the  elimination  causes  both  x*^ 
and  X  to  disappear,  thus  showing  the  equations  to  be  contra- 
clictory.     We  have  then  no  point  of  intersection. 

Ex.  2.     Consider 

4x-'_()y2_8x  — ,32  =  0, 
2x-  :ly  —  2  =  0. 
Substituting  from  the  second  equation  in  the  first,  we  obtain 
-  36  =  0. 

This  shows  the  equations  to  be  contradictory,  and  the  loci  do  not  inter- 
sect.    (Fig.  25.) 


We  may  examine  these  cases  by  the  method  of  limits  as  we  did  the 
exceptional  case  of  §  17.  In  general,  if  we  eliminate  y  between  a  given 
equation  of  the  second  degree  and  a  given  equation  of  the  first  degree, 
we  obtain  a  quadratic  equation  of  the  form 

ax-  +  bx  +  c  =  0. 


The  roots  of  tliis  equation  are 


X2' 


-  b  -h  Vb'^ 

-4ac 

2a 

-  b  -  Vb-'^ 

—  4ac 

2c 


2a 


b  —  Vb^  —  4ac' 

2c 

b-f-  Vb2  -  4ac" 


40  PLANE    ANALYTIC    GEOMETRY. 

If,  now,  a  approaches  zero  as  a  limit,  X2  increases  without  limit  and 
xi  approaches  the  value  —  ,  . 

If,  then,  a  is  made  to  approach  zero  by  altering  the  position  of  one  of 
the  given  loci  (see  §  17,  fine  type),  we  may  say  that  the  case  in  which 
we  find  only  one  solution  of  a  quadratic  and  a  simple  equation,  repre- 
sents the  limiting  case  of  the  intersection  of  two  loci,  when  one  point  of 
intersection  is  indefinitely  remote.     (See  Fig.  24.) 

If  both  a  and  b  approach  zero  as  a  limit,  both  xi  and  X2  increase  with- 
out limit.  Hence  the  case  in  which  the  quadratic  and  the  simple  equa- 
tion are  contradictory,  represents  the  limiting  case  of  the  intersection  of 
two  loci,  when  both  points  of  intersection  have  receded  indefinitely  from 
the  origin.     (See  Fig.  25.) 

EXAMPLES. 

Plot  the  following  straight  lines  : 
1.  5x  +  7y  +  15  =  (). 

3.  3x  4"  5y  =  0. 

Plot  the  following  curves  of  the  second  order  : 

4.  x2  +  y2  =  49. 
6.  y2  =  8x. 

6.  25x2  +  4/-' =  100. 

7.  x^  —  y^  =  4. 

8.  xy  =  12. 

9.  x2  +  4x— 4y  +  4  =  0. 

10.  x-'+y2  — 4x  +  Gy  —  3  =  0. 

11.  9x2  ^  igy2  _|_  i^^  _  (54y  _  71  ^  0. 

12.  9x2  _  4y2  _  3Q^  _^  24y  -  3(3  =  0. 

Plot  the  following  curves  of  the  third  order : 


13.  y  =  x". 

14.  f  =  xl 

15.  y2  =  X  (x  —  3)2. 


LOCI.  41 


16.    y2=x  (x  +  2)  (x-3). 

^       x-  +  4a 

18.  y^  =  ^^^^. 

Plot  the  curve  of  the  sixth  order  : 

19.  16y-  =  4x*-x''. 

Plot  the  f  olio  win  ar  transcendental  curves 


20. 

y  =  cos  x. 

21. 

y  z=:  tan  X. 

22. 

y  ^  sin~'x. 

23. 

y  =  10\ 

ot  the  following  curv 

24. 

r  =  ae. 

25. 

r  =  a  sin  20. 

26. 

r  =  a  cos  30. 

27. 

r  =  a  (1  —  cos  0). 

28. 

r  =  a  tan  0. 

29. 

■    3^ 

r  =  a  snr  -• 

30. 

r  =  a-  cos  2^. 

31.  Find  the  equation  of  the  locus  of  a  point  of  which  the 
distance  from  the  axis  of  x  equals  4  times  the  distance  from 
the  axis  of  y.     Plot  the  curve. 

32.  Find  the  locus  of  a  point  wluch,  if  connected  with  the 
points  (—  1,  3)  and  (—  3,  —  3),  forms  a  triangle  of  the  same 
area  as  if  connected  with  the  points  (1,  2),  (2,  2). 

33.  A  point  moves  so  that  its  distances  from  the  two  fixed 
points  (2,  —  3)  and  (—  1,  4)  are  equal.  Find  the  equation  of 
the  locus  and  plot. 

34.  A  triangle  of  area  25  has  two  vertices  at  the  points 


42  PLANE    ANALYTIC    GEOMETRY. 

(5,  —  6)  and  ( —  3,  4).     Find  the  locus  of  the  third  vertex  and 
plot. 

35.  Find  the  equation  of  the  locus  of  a  point  5  units  from 
the  point  ( —  1,  —  2)  and  plot  the  locus. 

36.  A  point  moves  so  that  its  distance  from  the  axis  of  y 
equals  its  distance  from  the  point  (4,  0).  Find  the  equation 
of  its  locus  and  plot. 

37.  A  point  moves  so  that  its  distance  from  the  origin  is 
equal  to  the  slope  of  the  straight  line  joining  it  to  the  origin. 
Find  the  equation  of  the  locus  in  polar  coordinates  and  plot. 

38.  A  point  moves  so  that  its  distance  from  the  origin 
equals  a  constant  times  the  angle  which  the  line  joining  it  to 
the  origin  makes  with  the  initial  line.  Find  the  equation  of 
the  locus  in  polar  coordinates  and  plot. 

39.  A  point  moves  so  that  its  distance  from  the  axis  of  x 
is  1  its  distance  from  the  point  (0,  4).  Find  the  equation  of 
the  locus  and  plot. 

40.  A  point  so  moves  that  its  distance  from  the  point  (1,  3) 
is  to  its  distance  from  the  point  ( —  4,  1)  in  the  ratio  of  2  :  3. 
Find  the  equation  of  its  locus  and  plot. 

41.  A  point  so  moves  that  its  distance  from  the  axis  of  y  is 
to  its  distance  from  the  point  (3,  2)  in  the  constant  ratio  2  :  3. 
Find  the  equation  of  its  locus  and  plot. 

Find  the  points  of  intersection  of  the  following  loci,  noting 
points  of  tangency  if  such  exist  : 

42.  5x  — 7y  +  13  =  0,,    2x  +  fiy  +  3==0. 

43.  x  +  y  — 7  =  0,    2x  +  3y  +  2  =  0. 

44.  y-3x-3  =  0,    5y-2x-2  =  0. 

45.  3x  +  2y  =  0,    x  —  4y  =  0. 

46.  x2  +  y2-4x  +  6y-12  =  0,    2y  =  3x  +  3. 

47.  x2  +  y2  — 4x  +  6y  — 12  =  0,     x  —  3  =  0. 

48.  y2-10x  — 6y  — 31  =  0,    2y-10x  — 47  =  0. 

49.  3x'^  +  7x  +  2y  +  2  =  0,    x  +  y -3  =  0. 

50.  2x-4-3y-  =  l,    2x-3y  +  3  =  0. 


LOCI.  43 

51.  4x2  +  9y2=r36,    x  +  3y  =  5. 

53.  9x"  — 4y^  +  54x-lGy  +  29  =  0,    2y  —  3x  +  5  =  0. 

54.  4x'-'  — 25y-=100,    2x-5y  =  0. 

55.  xy  +  3x-2y-8  =  0,    y  +  3  =  0. 

56.  x2  +  y2=r41,     xy  =  20. 

57.  X'  +  y-  —  Gx  -  2y  -  15  =  0, 
9x-  +  9y'^  +  6x  -  Gy  -  27  =  0. 

^^'  25  +  9  ~^'    25      9-^- 

For  what  values  of  b  will  the  following  curves  be  tangent? 

59.  x-  +  y2  =  49,    y  =  3x  +  b. 

60.  5  +  4=1,    y  =  ;;x  +  b. 

61.  y2  =  8x,    y==3x  +  b. 

What  must  be  the  values  of  m  in  order  that  the  following 
loci  may  be  tangent  ? 

62.  x2+y2  =  4,     y  =  mx  +  3. 

63.  9x2-16y2  =  144,    y  =  mx  +  3. 

64.  x2  +  y2  =  25,     y  =  mx  +  3. 

4 

65.  Show  that  y  ^=  mx -| is  tangent  to  y-^16x  for  all 

values  of  m. 

(The  following  examples  are  to  be  solved  by  finding  the 
intersection  of  two  loci  each  of  which  contains  the  required 
point.) 

66.  Find  a  point  equidistant  from  the  points  (—  3,  4), 
(5,  3),  and  (2,  0). 

67.  A  triangle  of  area  10  is  so  constructed  that  two  vertices 
/  are  at  the  points  ( —  2,  2)  and  (0,  0),  while  the  third  vertex  is 

on  the  line  x  +  3y  =  0.     Find  the  third  vertex. 


44  PLANE   ANALYTIC    GEOMETRY. 

68.  Find  a  point  on  the  axis  of  x  which  is  equidistant  from 
(5,  7)  and  (3,  -  4). 

69.  Find  a  point  on  the  line  x  —  4y  +  7  =;  0,  the  distance 
of  which  from  the  axis  of  x  equals  its  distance  from  the  axis 
of  y. 

70.  Find  the  points  on  the  curve  x^  +  y^  =  25,  such  that 
the  slope  of  the  line  joining  each  of  them  to  the  origin  is  2. 

71.  Find  the  points  which  are  5  units  distant  from  (2,  3) 
and  4  units  distant  from  the  axis  of  y. 


CHAPTER    III. 

THE   STRAIGHT    LINE. 

20.  In  the  last  chapter,  we  have  seen  that,  corresponding 
to  every  single  equation  containing  x  and  y,  there  is  a  locus 
such  that  the  coordinates  of  every  point  on  the  locus  satisfy 
the  equation,  and  the  coordinates  of  any  point  not  on  the 
locus  do  not  satisfy  the  equation. 

As  the  straight  line  is  the  simplest  locus,  we  will  find  its 
equation  first.  Now  a  straight  line  is  determined,  if  it  is 
made  to  pass  through  two  points,  or  through  one  point  in  a 
given  direction.  According  as  we  determine  the  line  in  one 
way  or  another  we  shall  expect  to  get  different  forms  of  the 
equation.  In  the  first  three  forms  of  the  equation  Avliich  we 
shall  find,  and  which  may  be  regarded  as  the  fundamental 
forms  of  the  equation,  i.e.,  [9],  [10],  [11],  the  line  [9]  and 
[11]  is  determined  by  being  passed  through  a  given  point  in 
a  given  direction,  and  the  line  [10],  by  being  passed  through 
two  given  points. 

21.  Equation  of  the  Straight  Line  in  terms  of  its  Slope  and 
Intercept  on  the  Axis  of  y. 

Let  AB  (Fig.  26)  be  the  given  line,  having  the  slope  m, 
and  cutting  off  on  YY'  the  intercept  OB,  denoted  by  b.  Let 
P  (x,  y)  be  any  point  on  the  line.  From  P  draw  PR  parallel 
to  YY',  and  from  B  draw  BR  parallel  to  XX'. 

Then  BR  ==  x,  and  RP  =  y  -  b. 

RP 

Since,  by  Trigonometry,  ^-^=  m,  we  have,  by  substitution, 

y-b 

=  m, 

X 

or  y  =  mx  +  b.  [9] 


46 


PLANE    ANALYTIC    GEOMETRY. 
Y 


As  P  is  any  point  of  tlie  line,  [9]  is  an  equation  which  is 
satisfied  by  the  coordinates  of  every  point  of  the  line,  and  which 
may  be  shown  to  be  not  satisfied  by  the  coordinates  of  any 
point  not  on  the  line.  For  if  any  point  not  on  the  line  be 
joined  to  B,  the  slope  of  this  joining  line  will  not  be  m,  as  it 
must  be  if  the  coordinates  of  the  point  are  to  satisfy  the 
equation.     Therefore,  [9]  is  the  required  equation  of  the  line. 

As  the  line  has  been  drawn  in  Fig.  26,  m  and  b  are  both 
negative ;  but  if  the  line  is  drawn  in  a  different  direction,  or 
cuts  off  a  different  intercept  on  the  axis  of  y,  m  and  b  will 
have  different  values.  And  since  m  is  the  tangent  of  an 
angle  between  0°  and  180°  and  b  is  a  distance,  m  and  b  can 
have  all  possible  values  from  minus  infinity  to  plus  infinity. 

Ex.  Find  the  equation  of  the  straight  Ime  making  an  angle  of  30° 
with  the  axis  of  x,  and  cutting  off  an  intercept  5  on  the  axis  of  y.      Here 


m  =  tan  30°  =  — ^'  and  b  =  5 ; 

V3 


•.  the  required  equation  is  y  =  — =  +  5. 


THE    STRAIGHT    LINE. 


47 


In  equation  [9],  m  and  b  are  constant,  and  x  and  y  vary  as 
different  points  on  the  line  are  taken.  If  we  give  new  values 
to  m  and  b,  as  rrii  and  bj,  y  ^  rriix  +  bi  will  represent  a  new 
straight  line,  in  the  equation  of  which  x  and  y  alone  vary. 

Now  quantities  such  as  m  and  b,  which  are  constant  for  any 
given  curve,  but  may  be  changed  so  as  to  obtain  the  equation 
of  a  second  curve  of  the  same  kind  as  the  first,  are  called 
parameters.  It  follows,  then,  that  the  parameters  do  not 
determine  the  kind  of  curve,  but  do  determine  which  one  of 
a  particular  kind  of  curves  has  been  selected. 

What  will  be  true  of  all  the  lines  for  which  b  =  0?  What  will  be 
true  of  all  the  lines  for  which  m  =  0  ?  If  the  equations  of  a  series  of 
lines  have  the  same  b,  but  different  m's,  what  is  true  of  all  the  lines? 
If  the  equations  of  a  series  of  lines  have  the  same  m,  but  different  b's, 
what  is  true  of  all  the  lines  ?  What  is  true  of  the  line  for  which  m  is 
infinite  ?     What  will  be  the  form  of  equation  of  this  last  line  ? 

22.  Equation  of  the  Straight  Line  in  terms  of  its  Inter- 
cepts on  the  Axes.  y 


Y 

Fig.  27. 


Let  AB  be  the  given  line,  cutting  the  axes  of  x  and  y  at 
points  A  and  B  respectively.     Denote  the  intercepts  OA  and 


48  PLANE    ANALYTIC    GEOMETRY. 

OB  by  a  and  b  respectively.     Let  P  (x,  y)  be  any  point  on  the 
line  and  draw  PM  parallel  to  YY'. 

Then  AM  =  x  —  a.  M P  =  y,  AO  =  —  a,  and  OB  =  b. 

The  triangles  AOB  and  AMP  are  similar,  since  their  sides 
are  respectively  parallel,  and  hence  their  homologous  sides  are 
proportional ; 

.  0B_  MP 

•"■  A0~  AM 
whence,  by  substitution, 

b  ^    y 

—  a       X  —  a' 

By  reduction,  we  get  the  required  equation, 

1+1='  M 

Ex.  1.  Find  the  equation  of  the  straight  line  making  intercepts  3  and 
5  on  the  axes  of  x  and  y  respectively. 

X         V 

Here,  a  =  3  and  b  =  5,  so  that  the  equation  is  ^^  +  r  =  L 

o      o 

Ex.  2.     Find  the  equation  of  the  straight  line  making  intercepts  3  and 

—  5  on  the  axes  of  x  and  y  respectively. 

Here,  a  =  3  and  b  =  —  5,  so  that  tlie  equation  is 

Q  +       r  =  1     or    -  -  i  =  L 
6       —  0  do 

In  equation  [10],  it  will  be  observed  that  a  and  b  are  the 
parameters,  since  they  are  constant  in  the  equation  of  any 
given  line,  and,  when  varied,  will  give  a  new  line  determined 
in  the  same  way  as  the  previous  lines  of  the  series  were 
determined. 

If  a  remains  constant  and  b  varies,  what  is  true  of  the  resulting  sys- 
tem of  lines  ?  If  a  and  b  vary  in  tlie  same  ratio,  what  is  true  of  the 
resulting  system  of  lines  ?  How  is  the  line  situated  for  which  a  is  finite 
and  b  infinite  ?  Througli  what  quadrants  in  the  plane  will  the  line  pass, 
for  which  a  and  b  are  botli  positive ;  for  which  a  is  negative  and  b  is 
positive  ?  "What  will  be  the  values  of  a  and  b  for  the  line  which  passes 
through  the  origin  ? 


THE    STRAIGHT    LINE. 


49 


23.  Equation  of  the  Straight  Line  in  terms  of  its  Normal 
Distance  from  the  Origin  and  the  Angle  the  Normal  makes 
with  the  Axis  of  x.  /t)u  ■ 


C.«<r-5^.  ci^ 


Fic.  28? 

Let  AB  be  the  given  line,  drawn  through  D  perpendicular 
to  OD.  Then  OD  will  be  the  normal  to  the  line  from  the 
origin,  and  the  length  of  OD,  denoted  by  p,  will  be  the  normal 
distance  of  the  line  from  the  origin.  Denote  the  angle  XOD 
by  a. 

Let  P  (x,  y)  be  any  point  of  the  line.  Draw  PM  parallel  to 
YY',  MR  parallel  to  AB,  and  PN  parallel  to  OD.  ,  , 

Then         Z  N  M  P  =  a, 

OR  =  OM  cos  a  =  X  cos  a, 

RD=  NP=  MP  sina^y  sin  a. 
But  OR+RD  =  OD; 

.  ■ .  X  cos  a  +  y  sin  a  =  p 
is  our  required  equation. 

This  form  of  tlie  equation  of  the  straight  line  is  called  the 
normal  equation  of  the  straight  line,  and  p  and  a  are  the  para- 
meters, p  having  all  values  from  0  to  oo,  and  a  having  all 
values  from  0°  to  360°. 


^\^^   \ 


fi^-^  61^. 


"=.     ^vw 


^   (U^^ 


-^    (Ur^  V^ 


wi  n 


50  PLANE    ANALYTIC    GEOMETRY. 

Through  what  point  will  the  line  pass  if  p  ==  0  ?  What  will  be  true  of 
the  system  of  lines  found  by  giving  p  dilferent  values  and  keeping  a  con- 
stant ?  What  will  be  true  of  the  system  of  lines  found  by  varying  a  and 
keeping  p  constant  ?  What  will  be  true  of  the  line  for  which  a  =  0° ; 
for  which  a  =  1)0°  ?  What  will  be  the  position  of  the  line  for  which 
a-0°  and  p  =  0  ? 

Note.  Equations  [9],  [10],  [11]  are  entirely  general  and  will  be  true 
of  the  line  determined  according  to  the  respective  conditions  imposed, 
in  whatever  position  it  may  be  drawn  in  the  plane ;  and  it  is  suggested 
that  the  student  prove  the  various  forms  of  equation  for  as  many  differ- 
ent positions  of  the  line  and  the  variable  point  on  the  line  as  he  can 
imagine,  by  constructing  new  figures  and  deriving  the  equations  from 
them,  as  was  done  in  §  3  of  Chap.  I. 

24.  Any  Equation  of  the  First  Degree  represents  a  Straight 
Line. 

All  the  forms  of  the  equation  of  the  straight  line  which  we 
have  found  are  of  the  first  degree.  This  fact  is  not  due  to 
the  way  in  which  the  lines  have  been  determined,  as  we  will 
show  by  proving  that  every  equation  of  the  first  degree  rep- 
resents a  straight  line. 

The  most  general  equation  of  the  first  degree  may  be  written 

in  the  form, 

Ax-|-By  +  C  =  0, 

where  A,  B,  and  C  are  arbitrary  constants. 

Let  Pi  (xi,  yi),  P2  (xo,  y^),  P3  (xa,  Ys) 

be  any  three  points  of  the  locus  of  this  equation, 

Ax+By  +  C==0. 

Since  the  coordinates  of  any  point  upon  the  locus  satisfy 
the  equation  of  the  locus, 

Ax,+  By,  +  C  =  0,  (1) 

Ax,,  +  Byo  +  C  =  0,  (2) 

Ax3+By,  +  C  =  0.  (3) 


THE    STRAIGHT    LINE. 


51 


By  subtracting  (2)  from  (1),  we  get 

A(x,-x,)  +  B  (y,-y,)  =  0, 

Xi  —  X2  B 

Similarly,  from  (1)  and  (3)  we  get 

yi  —  ya  ^  _  A  _ 
xi  —  xa  B 

Comparing  (4)  and  (5),  we  have 

yi~y2^yi-y3. 

Xi         Xj  Xi         Xg 

Y 


(4) 
(5) 
(6) 


P, 

^.^ 

— jN^ 

p,^^^.^^.- 

0      1 

-HN3 

Ms 

M, 

M, 

Y 

Fia.  29. 

Referring   to    Fig.   29,    we   see    that  (G)   is   the  same   as 

N2Pl^N3Pl 

PoN,       P,N3 

Hence  the  triangles  PsNoPi  and  P3N3P1  are  similar  by 
Plane  Geometry,  and  the  points  Pi,  Po,  and  Pj  must  be  on  the 
same  straight  line.  But  these  are  any  three  points  of  the 
locus  ;  therefore  all  points  of  the  locus  are  on  the  same 
straight  line,  and  hence  any  equation  of  the  first  degree 
represents  a  straight  line. 


52  PLANE    ANALYTIC    GEOMETRY. 

Moreover,  as  we  have  made  no  statement  dependent  for  its 
truth  on  the  axes  being  at  right  angles  to  each  other, 

Ax+  By  +  C  =  0 

represents  a  straight  line,  whether  the  axes  are  rectangular 
or  oblique,  while  formulas  [9]  and  [11]  have  been  proved  to 
hold  for  rectangular  coordinates  only. 

25.     Lemma. 

Any  two  quantities  I  and  n  are  proportional,  respectively,  to 
the  cosine  and  the  sine  of  some  angle. 

For,  if  — L^-A., 

cos  d      sin  6 

we  may  assume  the  value  of  this  ratio  to  be  r. 

Then  I  =  r  cos  6,  and  n  =  r  sin  6. 

Squaring  and  adding  these  equations,  we  have  r^=  l'^-)-  n"^, 

since  cos^  d  +  sin^  &=^^,  or  r  =  V  1'^  +  n^,  where  either  sign 

may  be  taken.     Hence 

I 
cos  6  =  ,  sm  6  = 


VP  +  n^  VI-  +  n^ 

26.  Determination  of  the  Parameters  of  any  Straight 
Line. 

It  is  obvious  that  any  straight  line  is  completely  determined 
if  the  parameters  a  and  b,  m  and  b,  or  p  and  a  are  known  ; 
and,  furthermore,  that  every  straight  line  must  have  each  of 
these  parameters,  though  its  equation  is  given  in  terms  of  two 
only. 

We  will  now  show  how  to  determine  the  five  parameters 
of  any  line,  the  equation  of  which  may  be  regarded  as  being 
in  the  form  Ax  +  By  +  C  ^  0. 

(1)  To  find  a  and  b.  From  the  definition  of  a,  it  is  seen  that 
a  is  the  abscissa  of  that  point  of  the  line  for  which  y  =  0. 
Therefore,  if   we    substitute    y  =  0  and  solve   the   resulting 


THE    STRAIGHT    LINE.  53 

equation  for  x,  we  shall  get  the  value  of  a,  which  in  this  case 

C  C 

will  be  —  -r  •     Similarly,  if  x  =  0,  y  =  b,  and  we  find  b  =  —  =  • 

Note.  If  A  and  B  decrease  toward  the  limit  zero,  while  C  remains 
constant,  it  is  evident  that  a  and  b  increase  indefinitely  and  the  line 
becomes  infinitely  distant  from  the  origin.  But  as  A  and  B  decrease 
toward  the  limit  zero,  the  limiting  form  of  the  equation  is  C  =  0.  For 
this  reason,  the  equation  C  =  0  is  said  to  be  the  equation  of  a  straight 
line  at  infinity. 

This  statement  must  not  be  taken  literally,  but  nuist  be  interpreted 
in  accordance  with  the  explanation  just  given. 

(2)  To  find  m.      We  may  write  the  equation  in  the  form 

By  =  —  Ax  —  C,  and    divide    by    B,    obtaining   as    a    result 

AC 
y  =  —  -X  —  —  •     But  this  equation  is  in  the  form  y  =  mx  4~  b, 

D  D 

A  C 

so  that  m  =  —  —  and  b  ^  —  — ,  this  last  value  of  b  being  the 

D  D 

same  as  that  we  found  before. 

(3)  To  find  a  and  p.  We  may  divide  our  equation  through 
by  VA'-^+  B^,  thus  })utting  it  in  the  form 

A  ,  B  ■  ^         =a 

Va^+b^''     Va-^+B2^     Va^+b^ 

where  we  shall  take  the  sign  of  the  radical  so  that  the  last 
term  shall  be  negative. 

By  the  lemma  of  §  25,  the  coefficients  of  x  and  y  are  now, 
respectively,  the  cosine  and  the  sine  of  some  angle,  while  the 
last  term  is  a  negative  quantity,  so  that  we  may  assume  the 
equation  to  be  in  the  form 

X  cos  a  +  y  sin  a  —  p  =  0, 

1  A  .  B  ,  -C 

where  cos  u  =  =-,  sm  a  =  —====-,  and  p: 


VA-+B-'  VA-+B-  VA'-^+B- 

Hence  :   to  put  any  equation  of  the  straight  line  in  the  form 
X  cos  a  +  y  sin  a  —  p  =  0, 
transjiose  all  the  terms  to  the  lift-hand  side  of  the  equation,  and 


54  PLANE   ANALYTIC    GEOMETRY. 

divide  by  the  square  root  of  the  stmt  of  the  squares  of  the  coeffi- 
cients of  X  and  y,  the  sign  of  the  radical  being  so  taken  that  the 
last  term  is  ?iegative. 

The  parameter  m  is  found  directly,  with  no  possibility  of 
ambiguity  of  sign,  and  m  is  the  tangent  of  the  angle  0  which 
the  line  makes  with  the  axis  of  x.  Since  0  is  never  greater 
than  180°,  if  m  is  positive,  6  is  in  the  first  quadrant;  while, 
if  m  is  negative,  0  is  in  the  second  quadrant ;  a,  on  the  other 
hand,  may  have  any  value  from  0°  to  360°,  so  that  it  cannot 
be  determined  from  the  value  of  a  single  one  of  its  trigono- 
metric functions,  as  the  sine  for  example.  If  the  sine  is  posi- 
tive, for  example,  we  should  know  only  that  a  is  in  the  first  or 
the  second  quadrant ;  but  if  sin  a  and  cos  a  are  both  positive, 
a  must  be  in  the  first  quadrant,  as  that  is  the  only  quadrant 
in  which  both  the  sine  and  the  cosine  are  positive.  Hence  in 
determining  a  it  is  7iecessary  to  tise  both  the  sine  and  the  cosine. 

It  might  seem  advisable  to  memorize  the  values  of  a,  b,  m, 
etc.,  in  terms  of  A,  B,  and  C,  as  found  above  :  it  is  much 
better,  however,  to  regard  the  work  of  their  derivation  only 
as  illustrative  of  the  methods  to  be  followed  in  any  similar 
problem. 

Ex.     Find  the  parameters  a,  b,  m,  p,  and  a-,  for  the  straight  line, 

2x -1-^-^2  =  0. 

Multiply  by  3  ;  Gx  +  y  -f  6  =  0. 

If  y  =  0,  Ox  -h  6  =  0,  whence  a  =  —  1. 

If  X  =  0,  y  -}-  6  =  0,  whence  b  ==  —  6. 

y  =  —  Gx  —  6  ;   .  •.  m  =  —  6. 

^  ,1,6 

—  x-f  — i=-y  +  — -F-  =0 


-  V37  -  V37  -  V37 


is  the  equation  after  dividing  through  by  —  VA-  -|-  B^. 

6        .  1  6 

.  •.  cos  a= =^1  sni  a— ;^)  p  =  — =z. 

V37  V37  V37 


THE   STRAIGHT    LINE.  65 

27.  Equations  connecting  the  Parameters  of  the  Straight 
Line. 

In  the  last  article,  we  have  shoAvn  how  to  tiiul  the  param- 
eters of  any  straight  line,  the  equation  of  which  is  given  in 
any  form  whatever,  by  reducing  the  equation  to  appropriate 
forms  and  noting  its  coefficients.  By  using  this  same  method 
of  procedure,  we  can  lind  the  relations  among  the  parameters 
a,  b,  m,  p,  and  a,  for  it  is  not  only  true  that  every  line  has  these 
five  parameters,  but  it  is  also  true  that  there  are  certain  fixed 
relations  connecting  them. 

Accordingly,  we  will  assume 

yrrrmX-fb,  (1) 

-+r=l,  (2) 

a       b 

X  cos  a-\-  y  sin  u  =  p,  (3) 

to  be  the  equations  of  one  and  the  same  straight  line,  and  will 
treat  each  of  these  equations  by  the  methods  of  the  previous 
paragraphs. 

From  (1)  :  if  x  =  0,  y  =  b,  and  nothing  new  is  learned ;  if 

b 
y  =r  0,  mx  +  b  =  0,  whence  a  ^= —  — 


If  we  put  (1)  in  the  normal  form,  we  have 
mx  .  y  b 


=  0; 


Vl  -\-  rf\-  Vl  +  m^  Vl  +  m- 

Treating  equation  (2)  in  the  same  wa}^,  we  get 

b  b  .  a  ,  ab 

m  = ,  cos  tt  =  — "-  ,  sin  a  =    ,  and  p  =  — 

a  Va-+b2  Va'-^+b^  Va-'+b- 

From  equation  (o),  we  find 

a= — '- — ,    b  =    ■       ,  and  m  =  —  etna. 

cos  a  sin  a 


66  PLANE    ANALYTIC    GEOMETRY. 


These  results  can  be  tabulated  as 

follows  : 

a  =  -^=-P-. 

m       cos  a 

1                    p 

J,           b  =  — ma  =  -r — . 
*                                    sin  a 

b 

m  = =  —  ctn  a . 

a 

m 

b 

Vl  +  m- 

Va-^  +  b-^ 

1 

a 

Vl  +  m-       Va-  +  b^ 

b                   ab 

Vl  +m-       Va-^  +  b- 

These  results  may  also  be  obtained  by  a  consideration  of 
the  geometrical  figure  from  which  these  equations  may  be 
derived. 

28.  Equations  of  the  Straight  Line  in  Oblique  Coordi- 
nates. 

By  §24,  we  know  that  these  equations  will  all  be  of  the  first 
degree,  and  we  shall  find  them  to  differ  but  very  little  from 
the  corresponding  forms  of  the  equation  in  rectangular  coordi- 
nates. 

First,  we  will  obtain  the  equation  of  the  line  when  it  is 
determined  by  b  and  the  angle  6. 

Let  AB  be  the  given  line,  cutting  off  the  intercept  b  on  the 
y  axis  and  making  an  angle  0  with  the  x  axis,  the  angle  XOY 
between  the  axes  being  w. 

Let  P(x,  y)  be  any  point  on  the  line,  and  draw  BR  and  PR 
parallel  to  the  axes  of  x  and  y  respectively. 

Then 

Z  RBP  =  ^,  Z  BPR  =  (o  — ^,  RP  =  y-b,  and  BR  =  x. 


THE    STEAIGHT    LINE. 

y 


57 


Fig.  30, 


.*.  by  Trigonometry, 

y-b 


whence 


sin  6       sin  (w  —  6) 
sin  0 


or  if,  for  brevity,  we  denote 
we  have 


sin  (w  —  9) 
sin  6 


x  +  b; 
by  n, 


sin  (o)  —  6) 
y  =  nx  +  b,  (1) 

exactly  the  same  form  that  we  found  for  rectangular  coordi- 
nates, but  with  a  new  meaning  for  the  coefficient  of  x. 
As  we  found  the  equation 


a       b 


(2) 


by  using  the  properties  of  similar  triangles,  it  is  evident  that 
this  equation  will  not  be  changed  at  all  by  the  use  of  oblique 
coordinates. 

We  have  left,  then,  the  determination  of  the  normal  equa- 
tion in  oblique  coiirdinates. 


58 


PLANE   ANALYTIC    GEOMETRY. 

y 


Fig.  31. 


AB  shall  be  the  line,  OD  being  equal  to  p,  and  Z^  XOD 
being  a.  P(x,  y)  shall  be  any  point  of  the  line.  Draw  PM 
parallel  to  YY',  MR  parallel  to  AB,  and  FN  parallel  to  OD. 

Then  ZN PI 


o)  —  a, 


OR  =  OM  cos  a  =  X  COS  a, 

RD=  NP=  MP  COS  ((o  — a)==y  COS  (w  — a). 

X  COS  a  +  y  COS  (co  —  o.)  ^^  R- 


(3) 


It  should  be  noted,  in  passing,  that  [9],  [10],  [11]  are 
really  special  cases,  respectively,  of  (1),  (2),  and  (3)  of  this 
article,  found  by  letting  w  =  90°. 


29.     Lemma. 

^„  cos  a cos    (oj  —  a) 

"  ~r-  B         ' 

the  value  of  this  common  ratio  will  be 

zh  sin  w 


VA^+  B2  — 2ABcosa) 


THE    STRAIGHT    LINE.  69 


_-         ,    ,  COS  a        COS  (w  —  a) 

For,  let  r  =  -j-  = ^^ ^  ; 

then  cos  a  =  Ar,  cos  w  cos  a  +  sin  w  sin  a  =  Br, 


or  cos  0)  (Ar)  +  sin  w Vl  —  A'^r'-'  =  Br, 

whence  we  find  r  equal  to  the  value  given  above. 

30.     Equations  connecting  the  Parameters  of  the  Straight 
Line  in  Oblique  Coordinates. 

We  have  as  the  equations  of  the  straight  line,  in  oblique 
coordinates, 

y  =  nx  +  b,  (1) 

X  cos  tt  +  y  cos  (w  —  a)  ^  P-  (3) 

By  following  out  the  same  methods  as  were  used  in  §  27, 
we  can  obtain  the  following  table  of  results  : 


b  =  —  an 


n       cos  a 

P 


n  ^  — 


cos  a 


cos   (w  —  a) 

b cos  a 

a  cos   (w  —  a) 

n  sin  0)  b  sin 


cos  (w  —  a)  ^ 


p  = 


Vn'^  +  1  +  2n  COS  0)       Vb''^^- a^  — 2ba  cos 
sin  w  a  sin  w 


Vn2+l  +  2n  cos  0)       Vb^  +  a^  —  2ba  cos  w 

b  sin  (X)  ab  sin  w 

Vn-^  +  l+2ncosa>~  Vb'"'  +  a^  -  2ba  cos  w 


60  PLANE   ANALYTIC    GEOMETRY. 

31.     Equation  of  the  Straight  Line  in  Polar  Coordinates. 


Fig.  32. 

Let  AB  be  the  given  line,  OD,  equal  to  p,  be  the  normal 
distance  of  the  line  from  the  origin,  and  let  /_  XOD  =  a. 

Let  P(r,  G)  be  any  point  on  AB.  Then  Z  DOP  =  ^— a, 
and  our  required  equation  is 

rcos(e  — a)  =  p.  [12] 

It  is  to  be  noted  that  in  this  equation  0  is  variable  and 
has  no  relationship  to  the  angle  the  line  AB  makes  with 
the  initial  line  OX. 

What  is  true  of  the  line  under  the  following  conditions  :  a  =  0° ; 
a  =  0°  and  p  =  0 ;  a  =  90° ;  a  =  90°  and  p  =  0  ? 

What  angle  will  the  line  r  cos  (B  —  45°)  =  5  make  with  the  initial  line  ? 
What  will  be  its  intercept  on  the  initial  line  ? 

32.     Distance  of  a  Point  from  a  Straight  Line. 

AB  shall  be  the  given  line,  its  equation  being  written  in 
the  normal  form,  x  cos  a  +  y  sin  a  =  p. 

Through  the  origin  draw  the  line  CD  parallel  to  AB. 

Then  there  will  be  three  cases  to  be  considered,  according 
as  the  point,  the  distance  of  which  from  AB  is  required,  is 
above  AB,  between  AB  and  CD,  or  below  CD. 


THE    STKAIGHT    LINE. 
Y 


61 


N 

N 

V 

^xP, 

\^^^ 

^V 

'^N^ 

^\    "-X     ^ 

\      ">-. 

v'     '^v^^) 

**                                    N 

N                                     S 

^          /                     N,                             "^ 

X-           P2^\ 

^                  ^^V 

/O    -k.          \a 

X                           / 

-^                       N                          \ 

V 

^^^x  ''"^^^^  ^ 

"^s 

^. 

'-N 

"-.(3)         ^^ 

/ 

Fig.  33. 

Case  I.  Let  Pi(xi,  yi)  be  the  given  point,  its  distance  from 
AB  being  pi.  Tlie  normal  distance  from  the  origin  of  a  line 
through  Pi  parallel  to  AB  will  be  p  +  pi,  and  the  angle  the 
normal  makes  with  the  axis  of  x  will  be  a.      Therefore  its 

equation  Avill  be 

X  cos  a  4"  y  sin  a  =  p  +  pi.  (1) 

But  Pi  is  a  point  of  this  line,  so  that 

Xi  cos  a  +  yi  sin  a  ^  p  +  pi, 
whence  pi  ^=  x,  cos  a  +  yi  sin  u  —  p. 

Case  II.     Let    P2(x2,  yz)  be  the  given    point,  its   distance 

from  AB  being  p.,.     The  normal  distance  of  a  line  through  P., 

parallel  to  AB  will  be  p  —  p^,  and  the  angle  the  normal  makes 

with  the  axis  of  x  will  be  a.     Therefore  its  equation  will  be 

X  cos  a  +  y  sin  a  =  p  —  po.  (2) 

But  P2  is  a  point  of  this  line,  so  that 

X2  cos  a  +  yo  sin  a  ^  p  —  p.i, 
whence  —  po  =■  y...  cos  a  +  y^  sin  a  —  p. 


62  PLANE    ANALYTIC    GEOMETRY. 

Case  III.  Let  P3(x3,  y^)  he  the  given  point,  its  distance 
from  AB  being  pg.  The  normal  distance  of  a  line  tlirough  Pg 
parallel  to  AB  will  be  pg — p,  and  the  angle  the  normal 
makes  with  the  axis  of  x  will  be  180°  +  «• 

Therefore  its  equation  will  be 

X  cos  (180°  +  a)  +  y  sin  (180°  +  a)  ^  pg  -  p, 
or  X  cos  a  +  y  sin  a  =  p  —  Pg.  (3) 

But  Pg  is  a  point  of  this  line,  so  that 

Xg  cos  a  -}-  yg  sin  a  =  p  —  pg, 

whence  —  Ps  =  Xg  cos  a  +  yg  sin  a  —  p. 

From  the  results  of  these  three  cases  we  may  formulate  the 
following  working  rule  : 

Put  the  equation  of  the  line  i?i  the  form 

X  cos  a  +  y  sin  a  —  p  =^  0 

and  substitute  for  x  and  y  the  coordinates  of  the  2^oint.  The 
resulting  value  of  the  left-hand  side  will  he  the  distance  of 
the  point  from  the  line,  and  will  be  positive  if  the  point  and 
the  origin  are  on  opposite  sides  of  the  line,  and  ivill  be  negative 
if  the  p)oi7it  and  the  origin  are  on  the  same  side  of  the  line. 

Ex.     Find  the  distance  of  the  point  (2,  3)  from  the  line 

2x  -  y  +  2  =  0. 
Writing  the  equation  in  the  form  x  cos  a  +  y  sin  a  —  p  =  0,  we  have 

V5       V5       VS 

,y.             •      1    r  .                    2(2)    ,3           2    _         3 
. '.  the  requu-ed  distance  — ^  H ^ -p.  = ; 

■\lb      ■\ro      yJl         V5 

3 
i.e.,  the  point  is  distant  — =  from  the  line,  and  on  the  same  side  of  the 

V5 
line  as  the  origin. 


THE    STRAIGHT    LINE. 


63 


33,     Equation  of  a  Line  through  any  Given  Point  with  a 
Given  Slope. 


Fig.  34. 

Let  AB  be  the  required  line,  having  the  given  slope  m,  and 
passing  through  the  given  point  Pi  (xj,  yi).  The  equation  of 
AB  call,  therefore,  be  written  in  the  form 

y  =  mx  +  b,  (1) 

where  at  present  b  is  unknown. 

But  since  the  line  passes  through  Pj,  Xi  and  yi  satisfy 
equation  (1)  ; 

•■•yi  =  mxi+  b, 
whence  b  =  yi  —  mxi. 

Substituting  in  (1)  the  value  of  b  thus  found,  and  reducing, 
we  get  the  required  equation 

y  —  yi  =  m  (X  — Xi).  [13] 

Ex.  1.  Find  the  equation  of  tlie  line  through  tlie  point  (1,  —3) 
making  /_  45°  with  OX. 

Here  m  =  tan  45°  =1,   xj  =  1,  yi  =  —  ^5. 

...  y +  .3=  l(x-l)  or  y-  x  +  4  =  0 
is  tlie  eqviation  of  the  line. 


64 


PLANE    ANALYTIC    GEOMETKY. 


Ex.  2.  Find  the  equation  of  the  line  through  (—  1,  —  2)  parallel  to 
the  line  x  +  2y  —  1  =  0. 

The  required  line  will  obviously  have  the  same  slope  as  the  given  line. 
X  +  2y  —  1=0  may  be  transformed  to  the  equivalent  form, 


-1,1         ■        - 
y--^x  +  --      .-.  m-- 

and  the  equation  of  the  required  line  is 
1 


^,    XI  =-1,   yi 


2, 


y  +  2=  —  -(x  +  1)    or    2y  +  x  +  5  =  0. 


34.     Equation  of  a  Line  through  Two  Given  Points. 

Y 


Let  Pifxj,  yi)  and  P2(x2,  yg)  be  the  two  given  points,  and 
let  P(x,  y)  1)6  any  point  on  the  line  through  tliem. 

DraAv  PiMi,  PM.^,  and  PM  all  parallel  to  YY'  and  PjS  par- 
allel to  XX'.  Then  the  triangles  PjSP  and  P1RP2  are  similar, 
their  sides  being  resjiectively  parallel,  so  that  their  homol- 
ogous sides  are  proportional. 


RP. 


PiS 
PiR' 


THE    STRAIGHT    LINE.  65 


whence  after  the  signs  of  the  denominators  are  changed, 

y  —yi^x  —  x, 
Yi-yj    xi  — X2' 

which  is  the  required  equation  of  tlie  line. 


[14] 


Ex.     Find  the  equation  of  the  line  through  the  points  (1,  —  2)  and 
(-2,4). 
Letting  (1,  —  2)  be  (xi,  yi)  and  (—2,  4)  be  (xo,  ya),  we  get 
V  4-  2         X  —  I 

We  must  now  consider  some  special  cases  to  which  [14] 
does  not  apply,  since  the  construction  fails. 

(1)  If  y2  =^  yi,  the  denominator  on  the  left  hand  side  be- 
comes zero,  and  hence  the  left  hand  member  seems  to  become 
infinite.  But  if  ya  ^  yi,  two  points  of  the  line  are  at  the 
same  distance,  i.e.,  yi,  from  the  axis  of  x  and  hence  the  line 
is  parallel  to  that  axis  and  its  equation  is  y  =  yi. 

(2)  If  X2  =  Xi,  we  can  prove  in  the  same  way  that  the  equa- 
tion of  the  line  is  x  =  Xx. 

(3)  If  yo  ==  yi  and  Xj  =  Xj,  both  denominators  become  zero. 
But  in  this  case  there  will  be  no  line  determined,  as  the  two 
points  which  were  to  be  used  in  its  determination  are  coinci- 
dent, and  through  any  one  point  an  infinite  number  of  lines 
can  be  drawn.  Therefore  we  should  not  expect  to  find  any 
equation. 

35,     Angle  between  Two  Lines. 

I  and  II  (Fig.  36)  shall  be  two  lines,  making  angles  0^  and  O2 
respectively  with  XX',  and  f3  shall  be  one  of  the  angles  made 
by  them  at  their  point  of  intersection.  As  we  liave  taken  /?, 
it  will  be  the  internal  angle  of  the  triangle  formed  with  the 
axis  of  X  by  the  two  linos. 

.■./3  =  o,-e„ 

J  ^       ^        tan  61  —  tan  Oo 

and  tan  B  =  j—, — ^ . 

1  +  tau  di  tan  62 


66 


PLANE  ANALYTIC  GEOMETRY. 


But  tan  ^1  =  nrii  and  tan  d^  =  ma,  where  rrii  and  mj  are  the 
respective  slopes  of  I  and  II. 


tan  p 


mi  —  ma 
l  +  mims 


[15] 


In  applying  this  formula,  the  value  of  tan  ^  may  be  found 
to  be  negative.  There  are  two  cases  in  which  this  may 
occur : 

(1)  As  chosen,  /8  may  be  an  obtuse  angle,  when  its 
tangent  will  necessarily  be  negative.  In  this  case  /3'  is  the 
acute  angle  between  the  two  lines,  and  )8'  :=  180°  —  /3,  so  that 
tan  /?'  =  tan  (180°  —  ^8)  =  —  tan  /5.  We  can  now  say  that 
the  acute  angle  between  the  two  lines  is  the  angle  of  which 
the  tangent  is  numerically  equal  to  the  value  found  by  formula 
[15]. 

(2)  If  we  do  not  plot  the  lines,  ^  may  be  acute,  but  we 
may  have  written  ^  =  O2  —  Oi  when  we  should  have  written 
I3=0i  —  62,  thus  finding  tan  (—  /S)  instead  of  tan  /?.  But 
tan  ( —  )8)  =  —  tan  yS,  so  that  in  this  case,  also,  we  may 
take  for  the  acute  angle  between  the  two  lines  the  angle  of 


THE    STRAIGHT    LINE.  67 

which  the  tangent  is  numerically  equal  to  the  value  found  by 
formula  [15]. 

To  sum  up  :  ive  may  alwai/s  take  rrii  and  m^  fvotn  the  equa- 
tions, substitute  in  formula  [15],  and  take  the  numerical  value 
of  the  result  as  the  tangent  of  the  acute  angle  between  the  lines; 
but  ive  shall  have  to  plot  the  lines  if  ire  wish  to  know  icliich  is 
the  acute  and  whicli  is  the  obtuse  angle. 

Ex.  Find  the  acute  angle  between  the  two  lines  y  =  2x  +  1  and 
2y  +  3x  —  1  =  0. 

Putting  these  equations  in  the  form  y  =  mx  +  b,  we  have  y  =  2x  +  1 

o  1  o 

and  y  =  —  ;^x  +  -,  whence  mi  =  2  and  mo  =  —  ^* 

-(-1)  7 


.  •.  tan  /3  = 

1  + 

and  the  acute  angle  between  the  two  lines  is  tan 


K-l)     '    . 


If  the  lines  are  parallel  to  each  other,  tan  fi^=0,  and 

mi  =  m2.  [16] 

If  the  lines  are  perpendicular  to  each  other, 
tan  (3  =  tan  90°  =  co ; 
.•.  1  -|-  miiTia  =  0, 

and  mi  =  -— •  [17] 

rn2 

As  the  cases  of  parallelism  and  perpendicvdarity  are  the 
most  important,  we  will  now  show  how  to  determine  by 
inspection  of  their  equations  when  two  given  lines  are  par- 
allel or  perpendicular  to  each  other. 

Let  the  lines  be 

Ax+  By  +  C  =  0, 
A'x  +  B'y  +  C'  =  0. 


68  PLANE   ANALYTIC    GEOMETRY. 

These  equations  may  be  written 

A  C 


y  =  -R'^- 


B  B 

B'""       B' 


.-.  mi  =  —  -  and  m.  =  —  —  • 

A       A' 
.'.if  the  lines  are  parallel,  —  =  —,  or  the  ratio  of  the  coefficients 

of  X  and  y  in  the  two  equatio72s  is  the  same  ;  and  if  the  lines 

A  B' 

are  perpcndicidar  to  each  other,  — ^^ —  -r-,,  or  the  ratio  of  the 

coefficients  of  x  and  y  in  one  equation  is  the  negative  of  the  ratio 
of  the  coefficients  of  ^  and  x  in  the  other. 

36.     Equation  of  a  Line  through  a  Given  Point  Perpendicu- 
lar to  a  Given  Line. 

Let  Pi(xi,  yi)  be  the  given  point,  and  let  the  equation  of 
the  given  line  be  written  in  the  form 

y  =  mx  +  b,  (1) 

The  equation  of  a  line  through  Pj  with  slope  mi  is  by  [13] 

§  33, 

y  —  yi  =  noi  (x  —  xi).  (2) 

But  if  lines  (1)  and  (2)  are  perpendicular  to  each  other, 

1 

mi=: 

m 

.•.our  required  equation  is 

y-yi  =  -^(x-xi).  [18] 

Ex.     Find  the  equation  of  a  line  through  the  point  (—  1,  —  3),  and 
perpendicular  to  the  line  x  +  2y  —  1  =  0. 

X  +  2y  —  1  =  0  is  equivalent  to  y  =  —  -  x  +  -• 
.  •.  y  +  3  = (x  +  1)  or  y  —  2x  +  1  =  0  is  the  required  equation. 


THE   STRAIGHT    LINE. 


69 


37.     Problem. 

To  find  the  equations  of  the  two  lines  which  pass  tlirongh 
the  point  (2,  3)  and  make  an  angle  tan""^  (^)  Avith  the  line 
3x  +  y  -  3  =  0. 

Y 


Fig.  37. 


Let  AB  be  the  given  line,  and  let  lines  (1)  and  (2)  be  the 
lines  making  the  required  angle,  tan~^  (■^),  with  AB. 

Then,  if  we  denote  tan~^  (^)  by  /3,  and  the  angles  made  by 
lines  (1)  and  (2)  with  XX'  by  6i  and  6^  respectively,  we  have 
from  triangle  ADF,  6i  =  $-\-  ^,  the  exterior  angle  of  a  tri- 
angle being  the  sum  of  the  two  opposite  interior  angles,  and 
from  triangle  C^E,6.,  =  6—ji  since  6  =  6.,-\- (3.  (Z  ACE  =  /3.) 
tan  0  -\-  tan  /? 


■.tan  6i 


=  -1 


and 


1  —  tan  6  tan  (3 

tan  6  —  tan  /3  ^ 

^''''  ^'  ~  l  +  tan^tan;8~  '' 
the  equations  of  the  lines  are : 

(1)  y_3^_i  (x-2)  or  y  +  x-5  =  0, 

(2)  y  — 3  =  7  (x  — 2)  or  y  — 7x  +  ll  =0. 


70 


PLANE   ANALYTIC   GEOMETRY. 


On  reviewing  the  work  of  this  article,  we  see  that  any 
problem  of  this  kind  can  be  solved  as  follows  :  Draw  the 
given  line  and  the  required  lines  of  such  lengths  as  to  form 
two  triangles,  in  each  of  which  the  sides  are  respectively  the 
axis  of  X,  the  given  line,  and  one  of  tlie  required  lines.  Then, 
since  vertical  angles  are  equal  and  the  exterior  angle  of  a 
triangle  is  the  sum  of  the  two  opposite  interior  angles,  we 
can  always  express  the  new  angles,  6i  and  62,  in  terms  of 
0  and  (3,  and  hence  determine  the  slopes  of  the  new  lines. 

It  may  happen,  however,  that  we  can  construct  only  one, 
or,  perhaps,  neither  of  these  triangles.  These  special  cases 
will  occur  when  one  or  both  of  the  required  lines  pass  through 
the  point  of  intersection  of  the  given  line  and  the  axis  of  x, 
in  which  cases  one  or  both  of  the  triangles  will  be  reduced  to 
a  point.  An  examination  of  the  corresponding  figures  below 
shows  that  no  new  difficulty  arises. 


X       X 


Fig.  38. 


h  =  0-/3. 


0  +  (3, 

e-(3. 


38.     Equation  of  a  Line  through  the  Intersection  of  Two 
Given  Lines, 

Ax  +  By  +  C  =  0  (1)     and    A'x+B'y  +  C'  =  0  (2). 

If  these  two  lines  meet  at  point  (xi,  yi),  we  find  by  solving 
(1)  and  (2)  simultaneously  that 


THE    STRAIGHT   LINE.  71 

_  BC'-B'C  _  CA'-C'A  * 

'^  ~  AB'  -  A'B  ""''"^  y^  -  AB'  -  A'B- 

Having  found  this  point  of  intersection,  we  can  write  down 
the  equation  of  tlie  line  passing  through  it  by  one  of  the  for- 
mulas, [13],  [14],  [18],  according  to  the  other  condition  im- 
posed upon  tlie  line. 

Ex.  Find  the  equation  of  the  line  through  the  point  of  intersection 
of  the  lines  y  —  2x  =  5  and  y  =  ox  -f  6,  and  the  point  (1,  1). 

The  given  lines  are  found,  on  solving  the  equations,  to  intersect  at 
(-  1,  3). 

.-.  the  equation  of  the  new  line  will  be,  by  [14]  §  34, 

There  is  another  method  of  dealing  with  this  same  class  of 
problems  which  is,  mathematically  speaking,  much  more  ele- 
gant.    It  is  as  follows  : 

If  I  and  k  are  any  two  arbitrary  midtipliers,  independent 
of  X  and  y,  we  may  form  the  equation 

l(Ax  +  By  +  C)  +  k(A'x  +  B'y  +  C)  =  0,  (3) 

which  is  the  equation  of  a  straight  line,  since  it  is  an  equa- 
tion of  the  first  degree. 

If  (xj,  yi)  is  the  point  of  intersection  of  lines  (1)  and  (2), 

Axi  +  Byi  +  C  =  0  and  A'x,  +  B'yi  +  C  =  0, 

whence     l(  Ax^  +  Byi  +  C)  +  k(A'x,  +  B'y,  +  C;  =  0. 

Hence,  for  any  and  all  values  that  nwiy  be  assigned  to  I  and 
k,  (3)  is  the  equation  of  a  straight  line  passing  through  the 
point  of  intersection  of  lines  (1)  and  (2).  We  may  accord- 
ingly impose  any  condition  we  wish  upon  I  and  k.  wlicn  the 
line  will  be  completely  determined. 

*  It  is  to  be  noted  that,  if  the  two  lines  are  parallel,  A B'  —  A'B  =  0, 
and  no  values  of  xi  and  yi  can  be  foimd.  For  a  further  discussion  of  this 
case  see  §  17. 


72 


PLANE   ANALYTIC   GEOMETRY. 


Applying  this  method  to  the  problem  just  solved  by  the  other  method, 
we  have 

l(y  -  2x  -  5)  +  k(y  -  3x  -  6)  =  0, 
a  line  passing  through  the  point  of  intersection  of  the  lines 
y  —  2x  —  5  =  0  and  y  —  3x  —  6  =  0. 
As  the  line  is  to  pass  through  (1,  1), 

1(1  -  2  -  5)  +  k(l  -  3  -  6)  =  0   or   I  =  -  -  k. 

.-.  -  ^k(y-2x  -5)  +  k(y-  3x  -  6)  =  0, 

or  y  +  X  =  2  is  the  required  line. 

It  is  to  be  noted  that  by  this  method  there  is  no  need  of 
finding  the  point  of  intersection  of  the  given  lines,  and  that 
it  holds  even  if  the  two  lines  are  parallel,  in  which  case,  of 
course,  the  required  line  is  also  parallel  to  the  two. 

39.  Equations-  of  the  Bisectors  of  the  Angles  between  Two 
Lines. 

Y 


Fig.  39. 


Let  the  equations  of  the  given  lines  be  written  in  the  forms 
X  cos  ai  +  y  sin  ai  —  Pi  =  0,  I. 

X  cos  a2  +  y  sin  a,  —  p.^  ^  0.  II. 


THE   STRAIGHT    LINE.  73 

Let  (1)  and  (2)  be  the  two  bisectors,  (1)  being  called  the 
internal  bisector  since  it  passes  through  that  one  of  the  angles 
made  by  I  and  II  which  includes  the  origin,  and  (2)  being 
called  the  external  bisector  since  it  does  not  pass  through  the 
angle  including  the  origin. 

We  know  by  Plane  Geometry  that  every  point  in  the 
bisector  of  an  angle  is  equally  distant  from  the  sides  of  the 
angle. 

Therefore,  if  (x,  y)  is  any  point  of  bisector  (1),  we  have,  by 
§32, 

X  cos  ai  +  y  sin  aj  —  pi  =  x  cos  ao  +  y  sin  an,  —  p.i,     (1) 

since  any  point  of  this  line  is  either  on  the  same  side  of  both 
lines  as  the  origin  or  on  the  opposite  side  of  both  lines  from 
the  origin,  so  that  the  values  of  its  distances  from  the  two 
lines  would  both  be  negative  or  both  be  positive. 
If  (x,  y)  is  any  point  of  line  (2),  by  §  32, 

X  cos  ai  -]-  y  sin  ui  —  Pi  =  —  (x  cos  uo  +  y  sin  a^  —  p-^,      (2) 

since  any  point  on  this  line  is  on  the  same  side  of  one  line 
as  the  origin  and  on  the  opposite  side  of  the  other  line  from 
tlie  origin,  so  that  the  values  of  its  distances  from  the  two 
lines  will  be  each  the  negative  of  the  other. 

It  is  evident  that  the  bisectors  pass  through  the  point  of 
intersection  of  the  given  lines  ;  for,  if  we  take  I  =  1  and  k  = 
—  1  in  the  work  of  §  38,  we  get  (1),  and  if  we  take  I  =  1  and 
k  =  1,  we  get  (2). 

Ex.  Find  the  equations  of  the  bisectors  of  the  angles  made  by  the 
lines  X  +  oy  —  7  =  0  and  ox  +  y  +  2  =  0. 

We  must  rewrite  these  equations  in  the  form  x  cos  a  +  y  sin  a  —  p 
=  0,  i.e., 

X      ,    ^y^ 7_  _  3x y 2_  _  ^ 

VlO       VlO       VlO  '  VlO       VlO      VlO 

.  •.  the  internal  bisector  is 

^     J.    3y  7  3x  y  2  .      I    <         r  _  n 

-=  +  —^ —  = -;_ —  or  4x  +  4y  -  5  -  0; 

VlO       VlO       VlO  VlO        VlO       VlO 


74  PLANE    ANALYTIC    GEOMETRY. 

and  the  external  bisector  is 

vio     Vio     Vio        V     Vio      Vio     Vio/ 

40.  Equations  of  Higher  Degree  than  the  First  Repre- 
senting Straight  Lines. 

We  have  seen,  iu  our  work  up  to  this  point,  that  every  form 
of  the  equation  of  the  straight  line  is  of  the  first  degree,  and 
conversely,  that  every  equation  of  the  first  degree  represents 
a  straight  line.  It  may  happen,  however,  that  an  equation  of 
higher  degree  than  the  first  will  have  as  its  locus  two  or  more 
straight  lines. 

For  example,  the  equation 

Ax2  +  2  Hxy  +  By-  +  2  Gx  +  2  Fy  +  C  =  0 
may  have  been  formed  by  multiplying  out  the  left-hand  side 
of  the  equation 

(rx  +  sy  +  t)  (r'x  +  s'y  +  t')  =  0. 

Now  tliis  last  equation  is  satisfied  if  either  factor  is  zero, 
the  other  factor  remaining  finite.  But  if  x  and  y  have 
such  values  as  to  make  rx  +  sy  +  t  ^  0,  x  and  y  must  be  the 
coordinates  of  a  point  on  the  line  rx  +  sy  +  t  =  0.  Similarly, 
if  x  and  y  have  such  values  as  to  make  r'x  +  s'y  +  t'  =  0,  they 
must  be  the  coordinates  of  a  point  of  the  line  r'x  +  s'y  +  t'  =  0. 

Therefore  all  points  of  the  straight  lines  rx  +  sy  +  t  =  0 
and  r'x  +  s'y  H~  t'  ^=  ^  ^^^^  points  of  the  locus  of  the  equation 
(rx  +  sy  +  t)  (r'x  +  s'y  +  t')  ==  0. 

Moreover,  any  point  not  on  either  of  these  two  straight 
lines  will  have  coordinates  such  that  neither  factor  will  be 
made  zero  by  their  substitution  in  the  equation,  hence  the 
product  will  not  be  zero,  and  the  point  is  not  a  point  of  the 
locus. 

Therefore  the   locus  of  the  equation 

(rx  +  sy  +  t)  (r'x  +  s'y  +  t')=0 
is  the  pair  of  straight  lines  rx  +  sy  +  t  ==  0,  r'x  +  s'y  +  t'  =^  0. 


THE    STRAIGHT    LINE.  75 

It  follows  that 

Ax^  +  2  Hxy  +  By=^  +  2  Gx  +  2  Fy  +  C  =  0 

will  represent  a  pair  of  straight  lines,  if  the  left  hand  side  can 
be  written  as  the  product  of  two  linear  factors  ;  that  an  equa- 
tion of  the  third  degree  will  represent  three  straight  lines,  if 
the  left-hand  side  can  be  written  as  the  pi'oduct  of  three 
linear  factors,  etc.,  the  right-hand  side  being  zero  in  each  case, 
of  course. 

For  example, 

x2  —  xy -f-  5x  —  2y -F  6  =  0 

is  the  equation  of  the  pair  of  straight  lines  x  -f  2  =  0  and  x  —  y  -f-  3  =  0, 
since  it  can  be  written  in  the  form,  (x  -|-  2)  (x  —  y  -f  3)  =  0. 

41.     Condition  that   the  Equation  of  the  Second  Degree 
represents  Two  Straight  Lines. 

Following  the  usual  method  of  factoring  the  expression 

Ax2+2  Hxy+  By- +  2  Gx  +  2  Fy  +  C, 

by  eqiiating  it  to  zero  and  finding  the  value  of  either  x  or  y, 
in  terms  of  the  other  and  the  coefficients  of  the  expression, 
we  get,  if  we  solve  for  x. 


_-(Hy  +  G)±V(H^-AB)y^4-2y(HG-AF)  +  (G--CA) 
^_  _ 

That  we  may  be  able  to  factor  into  rational  factors,  the 
above  values  of  x  must  not  be  surd,  at  least  with  y  under  the 
radical  sign  ;  hence  the  expression  under  the  radical  sign  must 
be  a  perfect  square. 

The  condition  that 

(H^- AB)  y-  +  2y  (HG  -  AF)  +  (G"  -  CA) 
shall  be  a  perfect  square  is 

4(HG- AF)2-4  (H^- AB)  (G--CA)  =  0, 
or  ABC  +  2  FGH- AF^- BG^- CH-=:0. 


76  PLANE   ANALYTIC   GEOMETRY. 

This  is,  therefore,  the  necessary  condition  that 

Ax2  +  2  Hxy  +  By2  +  2  Gx  +  2  Fy  +  C  =  0 
shall  represent  two  straight  lines. 

If  the  above  condition  is  fulfilled  for  any  quadratic  equation,  it  may 
be  factored  into  two  linear  factors.  The  three  cases  which  may  occur 
are  as  follows : 

(1)  If  the  two  factors  are  distinct  and  the  coefficients  are  real,  the 
equation  represents  two  real  straight  lines. 

(2)  If  any  of  the  coefficients  of  the  factors  are  imaginary,  i.e.,  contain 
the  square  root  of  a  negative  expression,  there  is  no  real  locus  corre- 
sponding to  the  equation.  We  may  say,  from  analogy,  if  we  choose,  that 
the  equation  represents  tivo  imaginary  straight  lines.  It  is  a  remarkable 
fact  that  one  and  the  same  real  point  may  be  found  on  each  of  the  two 
lines. 

(.3)  If  the  two  factors  are  equal,  each  factor  represents  the  same 
straight  line.  We  sometimes  say,  in  this  case,  that  the  equation  repre- 
sents two  coincident  straight  lines. 

Similar  considerations  hold  in  the  case  of  an  equation  of  degree  higher 
than  the  second,  so  that,  when  an  equation  can  be  factored  into  linear 
factors,  the  line  corresponding  to  any  factor  may  be  real  or  imaginary. 

Note.  If  A  =  0,  the  above  method  of  factoring  is  impossible,  but  if 
we  had  solved  for  y,  instead  of  x,  in  the  course  of  the  above  proof,  we 
should  have  found  the  same  necessary  condition.  If,  however,  both  A 
and  B  are  zero,  the  above  type  of  proof  will  be  unavailable,  though  the 
condition  is  equally  true. 

In  this  case  the  general  equation  reduces  to 

2  Hxy4-2Gx-|-2  Fy+ C  =  0, 

and  it  is  evident  that,  if  the  left-hand  member  can  be  separated  into 
linear  factors,  these  factors  must  be  of  the  type  rx  -j-  t  and  s'y  +  t'. 
Multiplying  rx  +  t  by  s'y  +  t',  we  have 

rs'xy  ■+■  rt'x  +  s'ty  -f-  tt'. 
Therefore,  if 

2  Hxy-l-2Gx  +  2  Fy-F  C  =0 

represents  the  two  straight  lines 

rx  -I-  t  =  0  and  s'y  -|-  t'  =  0, 

rs'  =  2  H,   rt'  =  2  G,  s't  =  2  F,  and  tt'  =  C. 


THE    STRAIGHT    LINE.  77 

But  (ilL(lt:)_(ttX(rs:)^^ 

or  2FG-CH  =  0. 

Therefore,  2  FG  —  C  H  =  0  is  the  necessary  condition  that 

2  Hxy +  2Gx  +  2  Fy  +  C  =  0 

shall  represent  two  straight  lines. 

But  if  A  and  B  are  both  zero,  the  original  condition  reduces  to 

2  FGH-CH-i  =  0, 
or  2  FG-CH  =0, 

since  H  cannot  be  zero  at  the  same  time  with  A  and  B  if  we  are  to  have 
an  equation  of  the  second  degree. 

Hence  we  see  that  the  necessary  condition  first  found  is  the  necessary 
condition  in  all  the  cases  which  can  occur. 


EXAMPLES. 

1.  Find  the  equation  of  the  line  of  which  the  slope  is  5  and 
the  intercept  on  OY  is  3. 

2.  Find  the  equation  of  the  line  passing  through  the  point 
(0,  —  2)  and  making  an  angle  of  135°  with  OX. 

3.  Find  the  equation  of  a  line  making  an  angle  of  60° 
with  OX  and  cutting  off  an  intercept  —  1  on  OY. 

4.  A  line  making  a  zero  intercept  on  the  axis  of  y  makes 
an  angle  of  150°  with  the  axis  of  x.      What  is  its  equation  ? 

5.  A  line  making  a  zero  angle  with  OX  cuts  OY  at  a  point 
—  2  units  from  the  origin.     What  is  its  equation  ? 

6.  What  is  the  equation  of  the  line  of  which  the  intercepts 
on  the  axes  of  x  and  y  are  respectively  2  and  —  4  ? 

7.  What  is  the  equation  of  the  line  of  which  the  intercepts 
on  the  axes  of  x  and  y  are  respectively  —  3  and  —  5  ? 

8.  A  line  is  distant  5  units  from  the  origin,  and  its  normal 
makes  an  angle  of  60°  with  OX.     What  is  its  equation? 

9.  The  normal  to  a  line,  which  is  distant  3  units  from  the 
origin,  makes  the  acute  angle  tan-i|-  ^[^i^  ^^q  axis  of  x. 
What  is  the  equation  of  the  line  ? 


78  PLANE   ANALYTIC    GEOMETRY. 

10.  A  line  distant  2  units  from  the  origin  makes  an  angle 
of  45°  with  the  axis  of  x.     What  is  its  equation? 

11.  The  normal  to  a  line,  which  passes  through  the  origin, 
makes  an  angle  tan~|-  with  OX.  What  is  the  equation  of 
the  line  ? 

12.  The  normal  to  a  line,  distant  7  units  from  the  origin, 
makes  an  angle  of  180°  with  OX.  What  is  the  equation  of 
the  line  ? 

Determine  the  parameters  a,  b,  p,  m,  and  a  for  the  fol- 
lowing lines : 

13.  y  =  3x-l. 

14.  ^-y-1. 

2      5 

15.  ^X+^y-2:::.0. 

16.  2x  +  y  =  0. 

17.  x  +  3  =  0. 

18.  2y  — 1  =  0. 

19.  Find  the  equation  of  a  line  making  an  intercept  3  on 
OY  and  an  angle  of  30°  with  OX,  when  the  angle  XOY  is  60°. 

20.  Find  the  equation  of  a  line  making  an  intercept  —  2  on 
OY  and  an  angle  of  15°  with  OX,  when  the  angle  XOY  is 
45°. 

21.  Find  the  equation  of  a  line  making  a  zero  intercept  on 
OY  and  bisecting  the  angles  of  the  second  and  the  fourth 
quadrants,  when  the  angle  XOY  equals  45°. 

22.  The  normal  to  a  line  distant  2  units  from  the  origin 
makes  an  angle  of  45°  with  OX,  the  angle  XOY  being  60°. 
What  is  the  equation  of  the  line  ? 

23.  The  normal  to  a  line  distant  5  units  from  the  origin 
makes  an  angle  of  30°  with  OX,  the  angle  XOY  being  75°. 
What  is  the  equation  of  the  line  ? 

24.  Show  that  the  equation  of  the  straight  line  may  be 
written 


THE    STRAIGHT    LINE.  79 

X  COS  a  +  y  COS  (3=p, 
where  p  is  the  length  of  the  normal  from  the  origin  and  a 
and  ^  are  the  angles  which  this  normal  makes  with  the  axes 
of  X  and  y  respectively,  the  angle  between  the  axes  being  of 
any  magnitude  whatever. 

25.  Determine  tlie  parameters  of  the  line  3x  —  2y  +  1  =  0, 
the  angle  between  the  axes  being  30°. 

26.  The  normal  to  a  line  distant  3  units  from  the  origin 
makes  an  angle  of  75°  with  the  initial  line.  What  is  the 
polar  equation  of  the  line  ? 

27.  Find  the  polar  equation  of  a  line  distant  4  units  from 
the  origin  and  making  an  angle  of  37°  with  the  initial  line. 

28.  Find  the  distance  of  (2,  3)  from  the  line  y  =  6x  —  1. 
On  which  side  of  the  line  is  the  point  ? 

29.  How  far  distant  from  the  line  x  +  3y  +  7  =  0  is  the 
point  (1,  —  4),  and  on  which  side  of  the  line  is  it  ? 

30.  How  far  distant  from  the  line  2x  —  3y  =  0  is  the  point 
(1,5)? 

31.  Find  the  distance  of  the  point  of  intersection  of  the 
lines  y  —  2x  —  5  =  0  and  2y  —  x  —  7  —  0  from  the  line  y  +  3x 
+  1=0. 

32.  Find  the  locus  of  the  points  at  the  constant  distance  3 
from  the  line  3x  +  4y  —  5  ^  0. 

33.  Find  the  locus  of  the  points  distant  2  units  from  the 
line  2x  +  y  +  2  =  0. 

34.  Find  the  locus  of  the  points  equally  distant  from  the 
lines  x  +  3y  —  7  =  0  and  3x  —  y  +  1  =  0. 

35.  Find  the  equation  of  tlie  line  through  the  point  (—3, 
—  5)  parallel  to  the  line  4x  —  3y  +  1  =  0. 

36.  Find  the  equation  of  a  line  through  (2,-1)  parallel 
to  the  line  x  —  y  =  6.  Where  will  the  new  line  cut  the  axes 
of  x  and  y  ? 

37.  Find  the  equation  of  a  line  through  (—  1,  —  3)  parallel 
to  the  line  2x  +  5y  —  3  =  0.  How  far  is  each  line  from  the 
origin  ?     How  far  apart  are  they  ? 


80  PLANE    ANALYTIC    GEOMETRY. 

38.  How  far  apart  are  the  parallel  lines  3x  +  4y  —  10  =  0 
and  3x  +  4y  -  7  ==  0  ? 

39.  Find  the  equation  of  a  line  parallel  to  the  line  3x  — 
2y  +  3  =  0  and  bisecting  the  line  joining  (1,  —  2)  and  (3,  4). 

40.  Find  the  equation  of  the  line  passing  through  the 
points  (—  3,  —  5)  and  (0,  6). 

41.  Find  the  equation  of  the  line  through  the  points  (2,  —  3) 
and  (4,  1). 

42.  Find  the  equation  of  the  line  through  the  points  (—1,6) 
and  ( —  1,  —  3). 

43.  Find  the  equation  of  a  line  through  (—  5,  —  3)  parallel 
to  the  line  joining  (5,  2)  and  (—  2,  —  5). 

44.  Find  the  acute  angle  between  the  lines  4x  —  oy+'^=0 
and  X  +  3y  +  2  =:  0. 

45.  Find  the  acute  angle  between  the  lines  x  +  3y  —  1=0 
and  3x  +  y  -  3  =  0. 

46.  Find  the  equation  of  a  straight  line  passing  through 
the  point  (—  5,  7)  perpendicular  to  3x  —  5y  +  6  =  0. 

47.  Find  the  equation  of  the  line  through  the  origin  perpen- 
dicular to  the  line  6x  +  3y  —  7  =  0. 

48.  Find  the  equation  of  a  line  perpendicular  to  2x  —  3y 
=  7  and  bisecting  that  portion  of  it  which  is  included  between 
the  axes. 

49.  Find  the  equation  of  the  perpendicular  bisector  of  the 
line  joining  (—  2,  —  4)  and  (—  4,  3). 

50.  What  are  the  equations  of  the  lines  which  pass  through 
the  point  (—  1,  0)  making  an  angle  tan-i  i  with  the  line 
2x-5y  +  2==0. 

51.  What  are  the  equations  of  the  lines  which  pass  through 
the  point  (—  5,  4)  making  an  angle  tan"  ^  i  with  the  line  15x 
_  3y  +  10  =  0. 

52.  Find  the  equations  of  the  lines  which  pass  through 
(3,  —  1)  making  an  angle  tan"  ^  f  with  5x  —  6y  +  3  =  0. 

53.  Find  the  equations  of  the  lines  passing  through  (1,  1) 
and  making  an  angle  tan- '  f  with  4x  —  3y  +  4  =  0. 


THE    STRAIGHT    LINE.  81 

54.  Find  the  equations  of  the  lines  passing  through  the 
point  (3,  —  5)  and  making  an  angle  tan"  ^  f  with  the  line 
x  =  4. 

55.  Find  the  equation  of  a  line  through  the  origin  and  the 
point  of  intersection  of  the  lines 

y  -  2x  -  5  =  0  and  x  -  2y  +  5  =  0. 

56.  Find  the  equation  of  a  line  through  (5,  —  5)  and  the 
point  of  intersection  of  the  lines 

X  — 2y  +  3  =  0  and  3x  +  4y  +  1  =  0. 

57.  Find  the  equation  of  the  line  passing  through  the  point 
of  intersection  of  2x  +  5y  +  6^0  and  3y  —  2x  +  6  =  0,  and 
parallel  to  x  —  y  +  1  ;=  0. 

58.  Find  the  equation  of  the  line  through  the  point  of 
intersection  of  3y  —  2x  +  5  =^  0  and  4y  +  5x  +  6  =  0,  and  per- 
pendicular to  2x  +  7y  +  8  ^  0. 

59.  Find  the  equation  of  a  line  through  the  point  of  inter- 
section of 

y-7x  +  36  =  0,  (1) 

X  -  3y  -  8  =  0,  (2) 

and  parallel  to  the  line  4x  -}-  3y  =  0,  without  finding  the  point 
of  intersection  of  (1)  and  (2). 

60.  Find  the  equation  of  a  line  through  the  point  of  inter- 
section of 

2x  -  3y  -  5  =  0,  (1) 

3x  -  4y  -  6  =  0,  (2) 

and  perpendicular  to  x  +  y  -f  1  =  0,  without  finding  the  point 
of  intersection  of  (1)  and  (2). 

61.  Find  the  equation  of  a  line  through  the  point  of  inter- 
section of 

2x  -  5y  -  41  =  0,  (1) 

7y-3x  +  58  =  0,  (2) 

and  the  point  (—  2,  —  9),  Avithout  finding  the  point  of  inter- 
section of  (1)  and  (2). 


82  PLANE    ANALYTIC    GEOMETRY. 

62.  Find  the  equation  of  a  line  thvougli  the  point  of  inter- 
section of  X  +  y  +  5  =  0  and  2x  +  L'y  +  ''^  =  0  and  tlie  point 

63.  Find  the  equ.ation  of  tlie  line    joining   the   points   of 
intersection  of  the  lines 

2x  —  y  -  5  =  O;  )         ,    (  2x  +  y  —  13  =  0, 
^  '    and    '  ^ 


3x  +  y-5  =  0,  j  (3x-y-2    =0. 

64.  Given  the  four  straight  lines, 

x-2y  +  2  =  0,  x  +  2y-2  =  0,  3x-y-3  =  0,  x  +  y+()  =  0, 

find  the  equations  of  the  three  other  lines,  each  of  which  con- 
tains two  of  the  six  points  of  intersection  of  the  given  lines. 

65.  Determine  the  value  of  m  so  that  the  line  y  :=  mx  +  3 
shall  pass  through  the  point  of  intersection  of  the  lines 
y  =  X  +  1  and  y  =  2x  +  2. 

66.  What  are  the  equations  of  the  bisectors  of  the  angles 
between  the  lines  3x  —  4y  +  T  =  0  and  12x  +  5y  —  6  =  0? 

67.  What  are  the  equations  of  the  bisectors  of  the  angles 
between  the  lines  V3x  -1-  V2y  =  0  and  x  +  2y  +  5=^0? 

68.  What  are  the  equations  of  the  bisectors  of  the  angles 
between  the  lines  3x  +  y  =  0  and  4x  —  3y  ^  0  ? 

69.  Find  the  bisectors  of  the  angles  between  the  lines  x  ==  0 
and  y  =  0. 

70.  Show  that  llx  +  3y  +  1  =  0  bisects  one  of  the  angles 
between  the  lines  12x  —  5y  +  7  =  0  and  3x  +  4y  —  2  =  0. 

71.  The  vertex  of  a  triangle  is  the  point  ( —  3,  —  4),  and 
the  base  is  the  line  joining  (2,  4)  and  ( —  1,  —  3).  Find  the 
lengths  of  the  base  and  the  altitude. 

72.  Find  the  vertices  and  the  angles  of  the  triangle,  tlie 
equations  of  the  sides  of  which  are  respectively 

y  =  0,    7x  — 3y  — 14  =  0,   and  x  +  y  —  12  =  0. 

The  vertices  of  a  triangle  ABC  are  the  points  ( — 3,  — 5), 
(2, -4),  and  (7,  6): 


THE    STRAIGHT    LINE.  83 

73.  Find  the  lengths  of  the  medial  lines  of  ABC. 

74.  Find  the  equations  of  the  sides  of  ABC. 

75.  Find  the  lengths  of  the  perpendiculars  drawn  from  the 
vertices  to  the  opposite  sides  of  ABC. 

76.  Find  the  equations  of  the  perpendicular  bisectors  of 
the  sides. 

77.  Where  do  these  bisectors  intersect  ? 

The  equations  of  the  sides  of  a  triangle  ABC  are  respec- 
tively 

y  +  4x  — 34  =  0,    3x  +  4y  — 19  =  0,    2y-5x  +  23  =  0: 

78.  Find  the  perimeter  and  the  area  of  ABC. 

79.  Find  the  equations  of  the  medians. 

80.  Find  the  point  of  intersection  of  the  medians. 

81.  Find  the  equations  of  the  lines  drawn  from  tlie  ver- 
tices perpendicular  to  the  opposite  sides. 

82.  Find  the    equations  of   the    lines  joining   the   middle 
points  of  the  adjacent  sides. 

83.  Find   the    simplest   form   of   the  polar  equation  of  a 
straight  line  through  the  origin. 

84.  What  locus  is  represented  by  the  equation  sin  3^  =:  0  ? 

85.  Prove  that  the  equation  of  the  line  through  the  two 
points  (ri,  Oi)  and  (r^,  62)  may  be  written 

r  [ri  sin  (O^  —  6)  —  r^  sin  {9.  —  6)~\  =  r^r.  sin  (^1  —  ^.,) 

or  more  symmetrically 

rri  sin  (6  —  0^)  +  x,r.  sin  {6^  —  6.^  +  r,r  sin  (Oo  —  6)=0. 

86.  Find  the  point  of  intersection  of  the  loci 


r  cos  (  ^  —  -  j  =  a,  r  cos  (  ^  —  -  1  =  a. 
87.  Find  the  points  of  intersection  of  the  loci 

r  cos  I  ^  ~  9  )  =  ~r  ''^"^  r  =  a  sin  d. 


84  PLANE   ANALYTIC    GEOMETRY. 

88.  One  diagonal  of  a  parallelogram  joins  the  points  (3, 
—  1)  and  ( —  2,  5).  One  extremity  of  tlie  other  diagonal  is 
( —  3,  —  4).     Find  its  equation  and  length. 

89.  Prove  by  calculation  that  the  perpendicular  bisectors  of 
the  sides  of  the  triangle  (2,  3),  (—  1,  —  2),  and  (—  3,  —  4) 
meet  in  a  point,  and  find  the  point. 

90.  Given  the  triangle  4x  —  3y  +  12  =  0,  3x  —  4y  —  6  =  0, 
12x  +  5y  +  60  =  0  ;  prove  by  calculation  that  the  bisector  of 
any  angle  divides  the  opposite  side  into  segments  proportional 
to  the  adjacent  sides. 

91.  Given  the  triangle  4x  —  3y  +  12  =  0,  3x  +  4y  —  6  =  0, 
12x  +  5y  +  30  :=  0  ;  prove  by  calculation  that  the  bisectors 
of  the  interior  angles  of  the  triangle  meet  in  a  point,  and  find 
the  point. 

92.  Prove  that  the  bisectors  of  the  angles  formed  by 
two  intersecting  straight  lines  are  perpendicular  to  each 
other. 

93.  Prove  analytically  that  the  locus  of  points  equally  dis- 
tant from  two  points  is  the  perpendicular  bisector  of  the  line 
joining  them. 

94.  Prove  analytically  that  the  medial  lines  of  a  triangle 
meet  in  a  point. 

95.  Prove  analytically  that  the  perpendicular  bisectors  of 
the  sides  of  a  triangle  meet  in  a  point.  (Take  one  vertex  of 
the  triangle  as  the  origin,  and  one  side  as  the  axis  of  x. 
Then  the  vertices  are  (0,  0),  (xi,  0),  and  (xg,  y2).) 

96.  Prove  analytically  that  the  perpendiculars  from  the 
vertices  of  a  triangle  to  the  opposite  sides  meet  in  a  point. 
(Take  axes  as  in  Ex.  95.) 

97.  Prove  analytically  that  the  perpendiculars  drawn  from 
any  two  vertices  to  the  median  from  the  third  vertex  are 
equal.     (Take  axes  as  in  Ex.  95.) 

98.  Prove  analytically  that  the  bisectors  of  the  interior 
and  the  exterior  angles  of  any  triangle  meet  by  threes  in  four 
points,   of   which  one  lies  inside  the  triangle.      (Take  the 


THE    STRAIGHT    LINE.  85 

origin   inside  the  triangle   and  write  the  equations  of   the 
sides  in  the  normal  form.) 

99.  Prove  analytically  that  the  lines  joining  the  middle 
points  of  the  opposite  sides  of  a  quadrilateral  bisect  each 
other. 

100.  Prove  analytically  that,  if  two  medians  of  a  triangle 
are  equal,  the  triangle  is  isosceles. 

101.  Show  that  the  locus  of  a  point  which  moves  so  that 
the  sum  of  its  distances  from  two  given  straight  lines  is  con- 
stant is  a  straight  line. 

102.  What  two  straight  lines  are  represented  by  the  equa- 
tion 2x2  +  3xy  +  y2  —  X  —  y  =  0  ? 

103.  What  two  straight  lines  are  represented  by  the  equa- 
tion 3x2  _^  8xy  —  3y2  +  6x  +  8y  +  3  =  0  ? 

104.  Plot  the  locus  of  the  equation  4x2  —  gy2  __  q 

105.  Plot  the  locus  of  the  eqviation 

2x2  _^  7xy  ^  3y2  _  X  _  3y  =  0. 

106.  Plot  the  locus  of  the  equation 

9x2  _^  9y2  _  6x  +  24y  -  19  ==  0. 

107.  Show  that  56x2  — 441  xy  —  56y2  — 79x  —  47y  +  9  =  0 
is  the  equation  of  the  bisectors  of  the  angles  between  the 
straight  lines  represented  by 

15x2  -  16xy  —  48y2  -  2x  +  16y  -  1  =  0. 

108.  Find  the  angle  between  the  two  straight  lines  repre- 
sented by  the  equation 

Ax2  +  2  Hxy+  By2==0. 

109.  Prove  that  any  homogeneous  equation  in  x  and  y  repre- 
sents a  system  of  straight  lines  passing  througli  the  origin. 


CHAPTER   IV. 

TRANSFORMATION    OF   COORDINATES. 

42.     Meaning  of  Transformation  of  Coordinates. 

So  far  in  our  work  we  have  dealt  with  the  Cartesian  coor- 
dinates of  any  point  in  the  plane  on  the  supposition  that  to  a 
given  point  corresponds  one,  and  only  one,  pair  of  coordinates ; 
and,  conversely,  that  to  any  pair  of  coordinates  corresponds 
one,  and  only  one,  point.  This  supposition  is  true  when  the 
axes  of  coordinates  are  once  fixed,  hut  a  moment's  thought 
will  make  it  evident  that  the  coordinates  of  the  point  are 
relative  quantities,  dependent  on  the  selection  of  the  axes, 
and  will  be  different  for  different  axes  chosen.  Moreover,  it 
is  obvious  that  the  axes  may  be  chosen  at  pleasure,  but  that 
there  must  be  some  relations  between  the  coordinates  of  a 
point  with  respect  to  one  set  of  axes  and  the  coordinates 
of  the  same  point  with  respect  to  a  second  set  of  axes. 

If  these  relations  are  known,  we  may  use  them  to  find  the 
coordinates  of  any  point  with  respect  to  a  new  set  of  axes,  its 
coordinates  being  given  with  respect  to  the  original  axes. 
We  are  then  said  to  make  a  trmisformation  of  coordinates,  and 
the  equations  expressing  the  relations  are  called  formulas  of 
transformatioii. 

It  must  be  borne  in  mind  that  a  transformation  of  coordi- 
nates never  alters  the  position  of  the  point  in  the  plane,  the 
coordinates  alone  being  changed  because  of  the  new  standards 
of  reference  adopted. 

Note.  In  all  the  work  that  follows,  OX'  and  OY'  should  be  carefully 
distinguished  from  the  OX'  and  OY'  used  up  to  this  point.  From  now 
on,  any  set  of  axes  will  be  lettered  only  at  the  origin  and  at  the  respec- 
tive positive  extremities  of  the  axes. 


TRANSFORMATION    OF    COORDINATES. 


87 


43.     Transformation  to  a  New  Origin,  the  New  Axes  being 
respectively  Parallel  to  the  Old  Axes. 

Y' 

,Y 


Let  OX  and  OY  be  the  original  axes,  and  O'X'  and  O'Y'  the 
new  axes  meeting  at  0',  the  coordinates  of  0'  with  respect  to 
the  original  axes  being  Xq  and  yo- 

Let  P  be  any  point  in  the  })lane,  its  coordinates  being  x  and 
y  with  respect  to  OX  and  OY,  and  x'  and  y'  with  respect  to 
O'X'  and  O'Y'.     Draw  PM'N  parallel  to  OY. 

Then  OM  ==  x,,,  MO'  =  yo, 

ON  =  x,  NP  =  y, 
0'M'  =  x',  M'P  =  y'. 

But  ON=OM  +  MN, 

NP=  NM'+  M'P; 

.•.x  =  Xo  +  x',   y  =  yu+y'-  [19] 

These  formulas  of  transformation  are  obviously  true 
whether  the  axes  are  oblicpie  or  rectangular. 

Ex.  1.  The  coordinates  of  a  certain  point  are  (2,  —  1).  What  will 
be  the  coordinates  of  this  same  point  with  respect  to  a  new  set  of  axes 
parallel  to  the  old,  and  meeting  at  the  point  (3,  —  2)  with  respect  to  OX 
and  OY  ? 


88 


PLANE    ANALYTIC    GEOMETRY. 


Here  xo  =  3,  yo  =  —  2,  x  =  2,  and  y  =  —  1. 

.-.  2  =  3  +  x'  and  —  1  =  —  2  +  y',  so  that  x'  =  —  1,  y'  =  1, 
Ex.  2.     Transform  the  equation 

f  +  4y  -  8x  +  28  =  0 
to  a  new  set  of  axes  parallel  to  the  original  axes  and  meeting  at  the  point 
(3,-2). 

The  formulas  of  transformation  are  x  =  3  -f-  x',  y  =  —  2  +  y',  by [19]. 
Therefore  our  equation  becomes 

(—  2  +  y')2  +  4(—  2  +  y')  -  8  (3  +  x')  +  28  =  0  or  y'^  -  8x'  -  0. 
As  no  point  of  the  curve  has  been  moved,  the  curve  is  changed  in  no  way 
whatever.     Its  equation  is  different  because  referred  to  neio  axes. 

After  the  work  of  a  transformation  has  been  completed,  the  primes 
may  be  dropped.  Therefore  the  equation  of  this  example  may  be  written 
y2  —  8x  =  0,  the  new  axes  being  now  the  only  ones  considered. 


44.     Transformation  from  one  Set  of  Rectangular  Axes  to 
another  Set  of  Rectangular  Axes  having  the  same  Origin,  and 
making  an  Angle  0  with  the  Original  Axes. 
Y 


Fig.  41. 


OX  and  OY  are  the  original  axes,  and  OX'  and  OY'  are  the 
new  axes  making  Z.  0  with  OX  and  OY  respectively.      Let  P 


TRANSFORMATION    OF    COORDINATES. 


89 


be  any  point  in  the  plane,  x  and  y  being  its  original  coordi- 
nates, x'  and  y'  its  new  coordinates. 

Draw  PM  parallel  to  OY,  PM'  parallel  to  OY',  M'N'  parallel 
to  OY,  and  M'N  parallel  to  OX.     Then  Z  NPM'  =  9,  and 
OM  ^x,  MP  =  y, 
OM'=x',  M'P  =  y'. 
But  OM  =  ON'- MN'=ON'- NM' 

=  OM'cos^—  M'Psin^, 
MP=MN+  NP=  N'M'+  NP 
=  0M'  sin  6+  M'P  cos  6. 

.'.  X  =  x'  COS  6  —  y'  sin  6, 
y  =  x'  sin  9  +  y'  cos  9. 

Ex.  Transform  the  equation  x  +  y  —  1  =  0  from  one  set  of  rect- 
angular axes  to  a  new  set  having  tlie  same  origin  and  bisecting  the 
angles  made  by  the  original  set. 

y' . 


[20] 


Here 


V2        V2'  V2        V2 


('^)^(^')— »»■■  — ^- 


0. 


45.     Transformation  from  a  Set  of  Rectangular  Axes  to  a 
Set  of  Oblique  Axes  having  the  Same  Origin. 
Y 


90 


PLANE    ANALYTIC    GEOMETKY. 


Let  OX  and  OY  be  the  original  axes  at  right  angles  to  each 
other,  and  OX'  and  OY'  be  the  new  axes,  ^^  XOX'  being 
denoted  by  6  and  /_  XOY'  by  6'.  The  oblique  angle  included 
by  the  new  axes  will  accordingly  be  6'  —  6. 

Let  P  be  any  point  in  the  plane,  its  original  coordinates 
being  x  and  y,  its  new,  x'  and  y'. 

Draw  PM  parallel  to  OY,  PM'  parallel  to  OY',  M'N'  parallel 
to  OY,  and  RM'N  parallel  to  OX. 

Then  ZRM'P  =  ^'. 

.•.OM  =  ON'+  N'M  =  ON'+  M'N  =  OM'  cos  0+  M'Pcos  6', 

MP=MN  +  N  P  =  N'M'+  NP  =  OM'  sin  6^  M'P  sin  9'. 

.'.  X  =  x'  cos  9  +  y'  cos  9', 
y  =  x'  sin  9  +  y'  sin  9'. 


[21] 


46.     Transformation  from  one  Set  of  Oblique  Axes  to  a  new 
Set  of  Oblique  Axes  with  the  Same  Origin. 

y 


Fig.  43. 


OX  and  OY  shall  be  the  original  axes,  Z  XOY  being  w. 
The  new  axes  shall  be  OX'  and  OY',  making  angles  6  and  & 
respectively  with  OX. 


TRANSFORMATION   OF   COORDINATES.  91 

Let  P  be  auy  point  in  the  plane,  its  original  coordinates 
being  x  and  y,  its  new,  x'  and  y'. 

Draw  PM  parallel  to  OY,  PM'  parallel  to  OY',  M'N'  parallel 
to  OY,  and  RM'N  parallel  to  OX.     Then 

ZN'M'0  =  w-^,  ZON'M'  =  180°-co,  ZM'NP  =  <o, 
ZRM'P=^',        ZM'PH=e'-io,       ZNM'P  =  180°-^'. 

.-.in  triangle  ON'M', 

ON'       _  OM'  _  N'M' 

sin  (<o  —  0)       sin  (180°  —  w)       sin  6 ' 

^,,,       OM'sin(cu  — ^)       x'sin(co  — ^) 
whence,     0N'= r-^= -— r^ ^, 


and  N'M'  = 


sm  o)  sm  u) 

OM'  sin  6      x'  sin  0 


sm  w  sm  (o 

since  sin  (180°  —  w)  =  sin  w. 
Also,  in  triangle  M'NP, 

NM'  M'P  NP 


sin  (0'  —  u>)      sin  <o      sin  (180°  —  $') ' 

„,,,       M'P  sin  (^'  —  o))       y'sm(6'  —  w) 

whence,     N  M '  = ^-^^ ^  = r^ '- 

sm  a>  sm  CO 


and 


M'P  sin  ^'      y'sin^' 

N  P  = -. =  -—. . 

sm  w  sm  w 

But  OM  =  ON'- MN'=ON'- NM', 

and  MP=MN  + NP=  N'M'+ NP. 

_  x^  sin  (ci)  —  9)  +  y^  sin  (o)  —  9^ 
sm  0) 
x'  sin  9  +  y'  sin  9' 

y  = ^ 

'  sm  0) 


[22] 


It  is  obvions  that  this  last  set  of  formulas  includes  all  the 
transformations  that  can  be  made,  when  the  origin  is  not 
changed,  different  values  being  assigned  to  w.  Q,  and  B\ 


92 


PLANE    ANALYTIC    GEOMETRY. 


47.     Double  Transformation  of  Coordinates. 

We  now  have,  on  tlie  one  hand,  a  formula  by  which  we  may 
transform  from  one  set  of  axes  to  a  new  set  of  axes  having  a 
new  origin  but  retaining  the  respective  directions  of  the  old 
axes,  and,  on  the  other  hand,  formulas  by  which  we  may  trans- 
form from  one  set  of  axes  to  a  new  set  of  axes  having  the 
same  origin.  But  it  often  happens  that  we  wish  to  transform 
to  a  new  set  of  axes  having  a  new  origin  and  not  parallel  to 
the  original  axes. 

Formulas  for  these  classes  of  transformations  may  be 
worked  out,  but  it  will  be  preferable  to  make  such  transfor- 
mations by  a  combination  of  the  formulas  already  found,  as 
shown  in  the  following  example. 

Transform  the  equation 

7x2+  i3y2_6V3xy-  (6V3-f-28)x+  (26  +  12  V3)y -|- 12  Vs -|- 25  =  0 

to  a  new  set  of  rectangular  axes,  making  an  angle  of  30°  with  the  original 
axes  and  meeting  at  the  point  (2,  —  1)  with  respect  to  the  original  axes. 


First  transform  to  the  new  axes  O'X"  and  O'Y"  meeting  at  0'  (2,  —  1) 
and  respectively  parallel  to  OX  and  OY. 


TRANSFOKMATION    OF    COORDINATES.  93 

By  [19],  §  43,  X  =  2  +  x",  y  =  -  1  +  y" 

will  be  the  formulas  of  transformation,  and  the  equation  becomes 
7(2+  x")2+13(-l+  y")2  — 6V3(2+  x")  (- 1  +  y") - 

(6V3  +  28)  (2+  x")+  (2G+  12  VS)  (-  1  +  y")  +  12V3  +  25  =  0, 
or  7  x"2  +  13y"2  -  6  ^/3x"y"  -  16  =  0. 

Now  by  formulas  [20],  §  44,  we  can  transform  to  the  required  axes 
O'X'  and  O'Y',  the  formulas  of  transformation  being 
„       x'Vs       y'      „       x'  ,    y'VS 

'^  =-^-2'y  =2+  ^• 

—  16  =  0  or  x'-  +  4y'2  -4  =  0. 

48.  In  reviewing  this  chapter,  we  see  that  the  expressions 
for  the  original  coordinates  in  terms  of  the  new  are  all  of  the 
first  degree,  so  that  the  result  of  any  transformation  cannot 
be  of  higher  degree  than  that  of  the  original  equation. 

On  the  other  hand,  the  result  cannot  be  of  lower  degree 
than  that  of  the  original  equation  ;  for  it  is  evident  that,  if 
we  transform  any  equation  to  a  new  set  of  axes  and  then  back 
to  the  original  set  of  axes,  the  equation  must  resume  its  origi- 
nal form  exactly.  Hence  if  the  degree  had  been  lowered  by 
the  first  transformation,  it  must  be  increased  to  its  original 
value  by  the  second  transformation ;  but  this  is  impossible 
as  we  have  just  noted. 

It  follows  that  the  result  of  any  transformation,  hence  of 
several  successive  transformations,  is  of  the  same  degree  as 
that  of  the  original  equation. 

Finally,  in  connection  with  this  chapter  the  student  should 
read  §  10,  on  transformation  from  rectangular  coordinates  to 
polar  coordinates,  and  conversely. 

EXAMPLES. 

1.  Transform  the  equation  3x  +  4y  +  14  =  0  to  a  new  set 
of  axes  parallel  to  the  original  axes  and  meeting  at  the  point 
(2,-5). 


94  PLANE    ANALYTIC    GEOMETRY. 

2.  Transform  the  equation  x"  +  y-  +  4x  —  6y  +  3  =  0  to  a 
new  set  of  axes  parallel  to  the  original  axes  and  meeting  at 
(-2,3). 

3.  Transform  the  equation  x"  +  12x^  —  y-  +  47x  —  2y -j- 59 
=  0  to  a  new  set  of  axes  parallel  to  the  original  axes  and 
meeting  at  ( —  4,  —  1). 

4.  Transform  the  equation  x^  —  2xy  +  y^  —  8  V2x  —  8  V2y 
=  0  to  a  new  set  of  axes  making  an  angle  of  45°  with  the 
original  axes  without  changing  the  origin. 

5.  Transform  the  equation  x^  —  y^  =  16  to  new  axes  bisect- 
ing the  angles  between  the  original  axes. 

6.  Show  that  the  equation  x"^  +  y^  =  a^  will  be  unchanged 
by  transformation  to  any  pair  of  rectangular  axes,  if  the 
origin  is  unchanged. 

7.  Transform  the  equation  x''  —  3y^  =  1  to  a  new  set  of  axes 
having  the  same  origin  as  the  original  axes,  the  new  axes  of 
X  and  y  making  the  angles  —  30°  and  30°  respectively  with 
the  original  axis  of  x. 

8.  What  will  be  the  equation  of  the  ciirve  represented  by 

X  V 

- — [-  — ^  1  in  rectangular  coordinates,  if  new  axes  of  x  and  y 

3  /      3 

are  taken  making  the  positive  angles  tan~^  -  and  tan^^  |  —  - 

respectively  with  the  original  axis  of  x,  the  origin  not  being 
changed  ? 

9.  After  transforming  the  axes  to  a  new  origin  (2,  —  1) 
and  then  rotating  them  through  an  angle  of  45°,  find  the 
resulting  form  of  the  equation 

x^  +  2xy  +y2  +  (4  V2  -  2)x  -  (2  +  4V2)y  +  (1  -  12  V2)  =  C- 

10.  Transform  the  equation 

3x2  _^  10  ^3xy  _  7y2  _  ^i^  _  30  V3)x  -  (42  +  30  V3)y 

-  (42  +  90  V3)  =  0 
to  a  new  set  of  axes  meeting  at  the  point  (3,  —  3)  and  making 
an  angle  of  30°  with  the  original  axes. 


TRANSFORMATION    OF    COORDINATES.  95 

11.  Wliat  will  be  the  equation  of  the  curve  represented 
by  3x-  +  4y'-  —  3x  +  4y  —  11  =  0,  if  new  axes,  parallel  to  the 
original  axes,  are  taken  so  that  there  shall  be  no  terms  of  the 
first  degree  in  the  transformed  equation  ? 

12.  Transform  the  equation  3x-'  +  6x-l-4y  —  5  =  0  to  a 
new  set  of  axes  parallel  to  the  original  axes,  so  choosing 
the  origin  that  there  shall  be  no  constant  term  and  no  term 
of  the  first  degree  in   x. 

13.  Transform  the  equation  3x-  +  Sxy  —  3y-^0  to  the 
straight  lines  it  represents,  as  axes. 

14.  The  equation  of  a  given  locus  is  7x^  —  2xy  +  4y'^^5, 
the  angle  between  the  axes  being  60°.  What  will  be  the 
equation  of  tliis  locus  if  the  axes  are  revolved  through  an 
angle  of  30°? 

15.  Prove  that  the  formulas  for  the  length  and  the  slope  of 
a  straight  line  and  for  the  area  of  a  triangle  are  not  changed 
by  a  transformation  to  a  new  set  of  axes  parallel  to  the  old. 

Transform  the  following  equations  to  polar  coordinates  : 

16.  X  cos  a  +  y  sin  a  =  p. 

17.  xy  =  7. 

18.  i+t=^- 

a-       b- 

19.  X-  +  y-  —  Say  —  Sax  =  0. 

20.  y-  (2a  —  x)  =  xl 

21.  x^+  x-y-  — ay-^O. 

Transform  the  following  equations  to  rectangular  coordi- 
nates : 

22.  r  cos  fo-'^\=- 10. 

23.  r  cos  fe-'^\-\-r  cos  f6-\-^,]  =  12. 

24.  r  =  a  sin  6. 

25.  r  =  a  (1  -  cos  6). 


96  PLANE    ANALYTIC    GEOMETRY. 

26.  r  =  a  sin  20. 

27.  r^  =  a'  cos  26. 

28.  r  =  a  (cos  2^  +  sin  20). 

29.  Prove  the  formula  for  distance  in  polar  coordinates  by 
transforming  to  polar  coordinates  the  same  formula  in  rect- 
angular coordinates. 

30.  Transform  to  polar  coordinates  the  formula  for  the 
area  of  a  triangle  in  rectangular  coordinates. 


CHAPTER   V. 

THE    CIRCLE. 

49.     Definition  and  Equation  of  the  Circle. 

Although  we  shall  see  later  on  that  the  circle  is  only  a 
special  case  of  a  more  general  class  of  curves  called  the  conic 
sections,  so  that  naturally  it  would  not  be  studied  by  itself  in 
detail  until  after  the  general  consideration  of  those  curves, 
nevertheless,  we  will  take  up  the  circle  here  for  the  following 
reason.  We  are  acquainted  with  the  circle  and  its  properties 
from  our  study  of  Plane  Geometry  ;  and  hence  for  us  it  is 
the  simplest  curve  in  the  plane,  and  the  easiest  on  which  to 
practice  our  new  methods,  lohicli  can  all  be  learned  in  this 
chapter. 

For  the  sake  of  learning  how  to  treat  a  curve  analytically 
and  of  acquiring  power  to  deal  with  new  curves  by  practice 
with  a  familiar  curve,  the  student  is  urged  not  to  solve  prob- 
lems by  his  knowledge  derived  from  Plane  Geometry,  if  he 
can  in  any  way  solve  them  analytically,  even  though  this 
requires  more  time. 

We  shall  adopt  the  usual  definition  of  a  circle,  i.e.. 

The  circle  is  the  locus  of  a  point  which  vioves  so  that  its  flis- 
tance  from  a  fixed  point,  called  the  centre,  shall  he  constant,  this 
constant  distance  being  called  the  radius. 

Let  P(x,  y),  Fig.  45,  be  any  point  of  the  circle  of  radius  r, 
of  which  the  centre  is  the  point  C(d,  e).  Then  by  [1],  §  3, 
the  equation  of  the  circle  will  be  , 

(x-di-'+(y->j^'^=r2.  [23] 

There  is  one  particular  f(jrm  of  this  equation,  which  is  of 
almost  as  much  importance  as  [23]  by  reason  of  the  uses  we 


98 


PLANE    ANALYTIC    GEOMETRY. 


can  make  of  it,  i.e.,  the  equation  of  the  circle  with  its  centre 
at  the  origin.     Here  d  =  e  =  0,  and  the  equation  is 

X'  +  y-  =  r\  [24] 


Fig.  45. 

I.  Find  the  equation  of  a  circle  tangent  to  OX.  2.  Find  the  equa- 
tion of  a  circle  tangent  to  both  OX  and  OY.  3.  Find  the  equation  of  a 
circle  tangent  to  OX  and  having  its  centre  on  OY.  4.  What  are  the 
parameters  of  a  circle  ?  5.  What  will  be  true  of  all  the  circles  having  the 
same  r,  but  different  d's  and  e's  ?  6.  What  will  be  true  of  all  the  circles 
having  the  same  d  and  e,  but  different  r's  ?  7.  All  circles  of  a  system 
have  the  same  d  and  r,  but  different  e's.     What  is  true  of  them  ? 


50.     General  Form  of  the  Equation  of  the  Circle. 

The  most  general  equation  of  the  second  degree  is 
Ax2  +  2  Hxy+  By2  +  2  $x -\-2^y -\- (^j=0. 


a) 


The  most  general  equation  of  the  circle,  when  expanded,  is 

x2  +  y2-2dx-2ey  +  (d?  +  e'-r2)  =  0.  (2) 

By  comparing  these  two  equations,  we  see  tliat  the  equation 

of  the  second  degree  represents  a  circle,  if  H  =:  0  and  B  =  A, 


THE    CIRCLE.  99 

i.e.,  if  no  xy  term  is  present  and  the  coefficients  of  x^  and  y^ 
are  equal,  for  then  we  can  divide  through  by  A  and  get  a  form 
exactly  like  (2). 

Therefore  Ax-  +  Ay"  -\-2  Gx-)-2  Fy  +  C  =  0  always  repre- 
sents a  circle  ;  or,  as  we  shall  prefer  to  write,  after  dividing 

through  by  A,  and  replacing^,  — ,  -~  by  a  new  G,  F,  and  C 
respectively,  i^ 

x^  +  y^  +  2  Gx  +  2  Fy  +  ^-  0  [25] 

always  represents  a  circle.  "  '^ 

This  equation  can  be  written 
(x2  +  2  Gx  +  G^)  +  (f  +  2  Fy  +F'^)--^G-'+  F'^  -  C, 
or  (x  +  G)'^  +  (y  +  F)'^  =  (  VG^  +  F^  -  C)'^ 

which  is  in  the  form  (x  —  d)"  +  (y  —  e)"  ^  r^. 
There  are  three  cases  which  can  occur  : 

(1)  If  G^  +  F^  —  O  0,  r  is  real,  and  we  have  a  real  circle. 

(2)  If  G^  +  F^  —  C  =  0,  r  =  0,  and  the  circle  reduces  to  a 
point,  the  coordinates  of  which  are  —  G  and  —  F. 

(3)  If  G^  +  F-  —  C  <C  0,  r  is  imaginary,  and  the  circle  is 
imaginary,  since  the  coordinates  of  no  real  point  satisfy  the 
equation. 

51.  Determination  of  the  Centre  and  the  Radius  of  a 
Circle  from  its  Equation. 

The  work  of  the  previous  article  suggests  a  method  of 
determining  the  centre  and  the  radius  of  a  circle  of  which  the 
equation  is  known,  as  illustrated  by  the  following  problem. 

Find  the  centre  and  the  radius  of  the  circle  4x2  4. 4y2_  i2x  +  4y  —  2  =  0. 
This  equation  may  be  put  into  form  [23]  as  follows  : 
Divide  by  4  :  x2  +  y2  —  3x  +  y  —  ^  =  0. 

Collect  terms  :  (x2  —  3x)  +  (y2  +  y)  =  1. 

Complete  squares  :    {x2  —  3x  +  |)  +  (y^  +  y  +  J)  =  J  +  }  +  \. 
Finally  :  (x  -  |)2  +  {y  +  \Y  =  Z. 

.-.  the  centre  is  at  the  point  (|,  — |),  and  the  radius  is  V3. 


100  PLANE    ANALYTIC    GEOMETRY. 

52.  Determination  of  the  Equation  of  a  Circle  Passing 
through  Three  Given  Points. 

We  know,  by  Plane  Geometry,  that  through  any  three 
points  not  in  the  same  straight  line  one,  and  only  one,  circle 
may  be  drawn.  In  Analytic  Geometry,  either  form  of  equa- 
tion of  the  circle  involves  three  parameters  (d,  e,  r,  or  G,  F,  C) 
which  require  for  their  determination  three  conditions,  and 
these  three  conditions  may  be  that  the  circle  shall  pass 
through  three  given  points. 

Assume,  then,  that  a  circle  is  to  pass  through  the  points 
Pi(Xb  yi)j  P2(x2)  ys) )  and  P3(x3,  ys).  Since  each  one  of  these 
points  is  on  the  curve,  their  coordinates  must  satisfy  the 
equation  of  the  curve.     Hence 

(Xl-d)^+(y,-e)~r^ 
(x2-d)^+(y,-e)--rS 
(x3-d)-^+(y3-e)-^-rl 

From  these  three  equations  we  can  find  the  values  of  d,  e, 
and  r,  and  on  substituting  them  in  the  original  equation  we 
shall  liave  the  equation  of  the  desired  circle. 

Ex.  Fiud  the  equation  of  a  circle  tlu'ough  the  tliree  points  (2,  —  2), 
(7,  3),  and  (6,  0). 

If  the  required  circle  is  (x  —  d)^  +  (y  —  e)^  =  r^, 
then  (2  -  d)2  +  (-  2  -  e)2  =  r2,  (1) 

(7  -  d)2  +  (.3  -  e)2  =  r2,  (2) 

(6  -  d)2  +  (0  -  6)2  =  r2.  (3) 

Solving  (1),  (2),  and  (3),  we  find  d  =  2,  e  =  3,  and  r  =  5.  Therefore, 
the  required  equation  is  (x  —  2)2  +  (y  —  3)2  =  25, 

or  x2  + y2  — 4x  —  Gy -12  =  0. 

Or  we  may  solve  this  class  of  problems  by  assuming  the 
equation  of  the  circle  to  be  in  the  form 

x2  +  y2  +  2  Gx  +  2Fy  +  C  =  0. 

If  the  circle  is  to  pass  through  the  points  Pi(xi,  yi), 
P2(x2,  y2),  and  P3(x3,  yg). 


THE   CIRCLE.  101 

xr  +  yr  +  2  Gxi  +  2  Fyi  +  C  =  0, 
X2'  +  y2'  +  2  Gx,  +  2  Fy,  +  C  =  0, 
X3-  +  y3-  +  2  GX3  +  2  Fy3+C  =  0. 

From  these  three  equations  we  may  determine  the  para- 
meters G,  F,  and  C,  and  thus  determine  the  equation  of  the 
circle. 

Applying  this  method  to  the  problem  just  solved,  we  have,  as  equations 
from  which  to  derive  G,  F,  and  C, 

4  +  4+    4G-4F+C  =  0,  or    4G-4F+C+    8  =  0, 
49  +  9  +  14  G  +  0  F  +  C  =  0,  "  14  G  +  (3  F  +  C  +  58  =  0, 
36  +  0  +  12  G  +  0  F  +  C  =  0,  "  12  G  +   C  +  30  =  0, 
whence  G  =  —2,  F  =  —  3,  and  C  =  —  12,  and  the  equation  of  the  circle 
is  x2  +  y2  +  2  (-  2)  X  +  2  (-  3)  y  +  (-  12)  =  0, 

or  x2  +  y-  —  4x  —  Gy  —  12  =  0. 

Note.  If  the  three  points  should  happen  to  be  on  the  same  straight 
line,  the  three  equations  to  be  solved  for  the  parameters  will  be  found 
to  be  inconsistent,  or  impossible  of  solution. 

53.  Determination  of  the  Tangent  to  a  Circle  when  its 
Slope  is  Given. 

Let  the  slope  of  the  required  tangent  be  m,  so  that  its 
equation  may  be  written  in  the  form  y  =  mx  +  b,  (1),  where 
m  is  given  and  b  is  to  be  so  determined  that  the  straight  line 
(1)  shall  be  tangent  to  the  circle.  We  will  take  the  equation 
of  the  circle  in  the  special  form,  x-4-  y^=  r^.     (2) 

If  (1)  is  tangent  to  (2),  the  straight  line  and  the  circle 
meet  in  two  coincident  points  according  to  §  18.  Substituting 
in  (2)  the  value  of  y  from  (1),  we  have 

x^  +  (nrix  +  b)-  =:  r^, 
or  (1  +  m-)  x2  +  2  mbx  +  (b-— r')=0. 

For  tangency  this  equation  must  have  equal  roots. 
.-.4  m-b^  —  4  (1  +  m-)  (b^  -  r^)  =  0, 
and  solving  this  equation  for  b  we  find 
b  =  ±  r  Vl  +  m^. 


102 


PLANE   ANALYTIC    GEOMETRY. 


Fig.  4G. 


The  double  sign  for  the  value  of  b  shows  that  there  are  two 
lines  with  the  given  slope,  Fig.  46,  each  of  which  cuts  the 
circle  in  two  coincident  points. 

.*.  y  ^  mx  +  r  Vl  -f  m-  (3) 

and  y  =  mx  —  r  Vl  +  m^  (4) 

are   both  tangents  to  the   circle    x^  -|-  y-  =  r^,  satisfying  the 
imposed  condition. 

The  above  method  of  finding  the  tangent  when  its  slope  is 
given  can  be  used  with  any  form  of  the  equation  of  the  circle, 
and  is,  moreover,  the  method  to  be  employed  in  all  numerical 
problems  of  this  kind. 

For  example :     Find  the  equation  of  a  tangent  to  the  circle 
x2  +  y2  _  4x  +  Gy  +  8  =  0, 
making  an  angle  tan-i  i  with  OX. 

Here  m  =  ^,  so  that  the  tangent  is  y  =  |x  +  b,  b  being  as  yet  undeter- 
mined. 

Substituting  the  value  of  y  from  this  equation  in  the  equation  of  the 
circle,  we  have      x2  +  (^x  +  b)2  — 4x  +  6  (^x  +  b)  +  8  =  0, 
or  5x-^  +  4  (b  -  1)  x  +  4  (b2  +  6  b  +  8)  =  0. 


THE    CIRCLE.  103 

For  tangency  this  equation  must  have  equal  roots,  so  that 
IG  (b  -  1)-^  -  4  (5)  ^4(b2  +  Gb+8)^  =0, 
whence  b  =  —  |  or  —  -\f. 
.:  the  two  tangents  are 

1         3  1  13 

or  2y  =  X  —  3,        2y  =  x  —  13. 

54.  Equation  of  a  Tangent  to  a  Circle  when  the  Point  of 
Contact  is  Given. 

We  shall  define  the  tangent  at  any  point  of  the  curve  as 
the  straight  line  which  the  secant,  through  that  point  and  any 
second  jjoint  of  the  curve,  apjjvoaches  as  a  limit,  as  the  second 
point  approaches  the  first  along  the  curve. 


Fig.  47. 

We  will  take  the  equation  of  the  circle  in  its  most  general 
form, 

x'^  +  y-  +  2  Gx  +  2  Fy  +  C  =  0.  (1) 

Pi(xi)  yO  shall  be  the  point  of  contact,  P2(x2,  ys)  any 
second  point  of  the  curve.  The  equation  of  the  secant  will 
be,  by  [14],  §  34, 

y-y,3.^^^(x-xO.  (2) 

Since  we  should  get  an  indeterminate  form  on  the  right-hand 
side  of  the  equation,  as  it  now  stands,  if  we  should  try  to  find 


104  PLANE   ANALYTIC    GEOMETRY. 

the  limiting  form  when  x.,  =  Xj  and  yj  =  yi,  we  must  find  an 

Vi  —  V" 
equivalent  expression  for ^,  the  factor  which  becomes 

Xj  X2 

indeterminate.  ^ 
Since  P^  and  P2  are  points  of  the  curve, 

Xi^+yr  +  2Gxi  +  2  Fyi  +  C-0,  (3) 

Xo'  +  yo^  +  2  GX2  +  2  Fy2+C  =  0.  (4) 
Subtract  (4)  from  (3)  : 
(xi^  -  X2^)  +  (yi^  -  y,')  +  2  G  (X,  -  X2)  +  2  F  (y^  -  y^)  -  0, 

or     (xx  — X2)(xi+x2  +  2  G)  +  (y,-y2)(yi  +  y.,  +  2  F)  =  0, 

1  yi  —  y-2        xi  +  xo  +  2  G  ,„, 

whence  x^^,  =  "  y, +-y;+-^ '  <«> 

.'.the  equation  of  the  secant  may  be  written 

X1  +  X2  +  2  G 
y-y^  =  -y,  +  y2  +  2F(^-^^)-  ^^^ 

Taking  the  limiting  form  of  this  equation  as  X2  and  ya 
approach  Xj  and  yi  respectively  as  limits,  we  have,  as  the 
equation  of  the  tangent  at  the  point  Pi(xi,  yj), 

2x,  +  2G,  . 

y-y^^-27rf2T(^-^^>' 

or  y-yi  =  -^i^(x-xO.  (7) 

Eedueing  (7)  we  get 

xjx  —  xi"  +  yiy  —  yi^  +  Gx  —  Gxi  +  Fy  —  Fyi  =  0.        (8) 
By  adding  (3)  to  (8)  we  transform  (8)  to 

Xix  +  yjy  +  G  (X  +  x^)  +  F  (y  +  y,)  +  C  =  0,      [26] 
as  the  form  of  equation  desired. 

Note.     If  the  equation  of  the  circle  be  written  in  the  form 

XX  4-  yy  +  G  (X  +  X)  +  F  (y  +  y)  +  C  =  0, 


THE   CIRCLE.  105 

it  will  be  noted  that  the  equation  of  the  tangent  at  any  point  (xi,  yi)  may 
be  written  by  replacing  one  of  the  two  x's  or  one  of  the  two  y's,  as  the 
case  may  be,  in  each  term  by  an  xi  and  yi  respectively. 

Ex.     Find  the  equation  of  the  tangent  to  the  circle 
x2  +  y2  _  4x  +  6y  +  8  =  0, 

passing  through  the  point  (1,-1). 

Since  (1,  —  1)  satisties  the  equation  of  the  circle,  it  is  the  point  of  con- 
tact, and  the  equation  of  the  tangent  is,  by  [26], 

Ix  +  (-  1)  y  +  {-  2)  (X  +  1)  +  3  (y  -  1)  +  8  =  0, 
or  X  —  2y  —  o  =  0. 

55.     Normal. 

The  normal  to  a  mime  at  any  point  is  the  straight  line  pass- 
ing through  the  jwint  and  j^&ipsndicular  to  the  tangeiit  at  that 
point. 

It  is  possible  to  find  the  formula  for  the  equation  of  a 
normal  to  a  curve,  but,  since  the  formula  is  not  symmetrical  as 
is  that  for  the  tangent,  it  is  not  easily  committed  to  memory. 
Accordingly,  in  the  solution  of  problems,  it  is  better  to  work 
out  the  equation  from  the  definition  above,  as  illustrated  in 
the  following  two  cases. 

Ex.  1.     Find  the  equation  of  the  normal  to  the  circle 

x2  +  y'-2-4x  +()y  +  8=0, 
at  the  point  (1,  —  1). 

In  §  54,  we  have  found  the  equation  of  the  tangent  at  this  point  to  be 

X  —  2y  —  .S  =  0. 
.-.  the  normal  will  be  y  +  1  = (x  —  1), 

y  +  2x  -  1  =  0. 

Ex.  2.     Find  the  normal  to  the  circle 

x2  +  y-2_4x +  6y +  8  =  0,  (1) 

corresponding  to  the  tangent  which  makes  an  angle  tan-'  |  with  the 
axis  of  X. 


106  PLANE   ANALYTIC   GEOMETRY. 

In  §  63,  we  found  one  of  tliese  tfingents  to  be 

X  —  2y  —  3  =  0.  (2) 

Solving  (1)  and  (2)  simultaneously,  we  find  the  point  of  contact  to  be 

(1,  —1).   Therefore,  the  normal  is  y  +  1  =  —  -  (x  —  1),  or  y  +  2x  —  1  =  0. 

2 

Every  normal  to  a  circle  passes  through  its  centre.     For  if 
the  equation  of  the  circle  is  in  the  form 

x^  +  y-  +  2  Gx  +  2  Fy  +  C  =  0, 
its   centre   is   ( —  G,  —  F),    and   the   tangent   at   any   point 
(xij  yO  is 

xix  +  yiy  +  G  (x  +  xi)  +  F  (y  +  yi)  +  C  =  0. 

Then,  by  [18],  §  36,  and  the  definition  of  the  normal,  the 
equation  of  the  normal  at  the  point  (xj,  yj)  will  be 

yi+F  .         . 
y-yx    _    x-x,    . 


yi-(-F)      x,-(-G) 

Since  this  last  is  obviously  the  equation  of  a  straight  line 
through  the  points  (xi,  yj)  and  (—  G,  —  F),  we  have  proved 
that  every  normal  to  a  circle  passes  through  the  centre  — 
every  normal,  for  (xi,  yj)  is  any  point  of  the  circle. 

Note.  This  property  of  the  normal  enables  us  to  solve  many  problems 
involving  normals  to  a  circle,  which  could  not  be  treated  by  either  of  the 
previous  methods,  such  as  finding  the  normal  passing  through  any  point 
not  on  the  circle,  etc. 

56.     Subtangent  and  Subnormal. 

The  subtangent  is  the  projection  upon  OX  of  that  portion 
of  the  tangent  included  between  the  point  of  tangency  and 
OX. 

The  subnormal  is  the  projection  upon  OX  of  that  portion 
of  the  normal  included  between  the  point  of  the  circle  and 
OX. 


THE   CIRCLE. 


107 


We  will  illustrate  these  quantities  and  their  evaluation  by 
the  following  numerical  example, 

Y 

\ 


Fig.  48. 


The  circle  is  x^  +  y2  =  25.     Tlie  coiJrdinates  of  P  are  3  and  4.     PT  is 
the  tangent,  OP  the  normal,  MT  the  subtangent,  and  OM  the  subnormal. 


The  equation  of  PT  is  3x  +  4y  =  25 
.-.  MT 


25  10 

3       '^~  3 


coordinates  of  T  are 
0M  =  3. 


and  0. 


57.     Length  of  a  Tangent  Drawn  from  Pi(xi,  yi)  to   the 
Circle  x^  +  y^  +  2  Gx  +  2  Fy  +  C  =  0. 


Fio.  49. 


108  PLANE    ANALYTIC    GEOMETRY. 

Draw  the  tangent  PiP,  also  CP  and  CPj,  C  being  the  centre 
of  the  circle.     Then  CPPi  is  a  right  triangle  and 

P^P=^CPi'-CP'. 
The  equation  of  the  circle  may  be  put  in  the  form 

(x  +  G)2+(y+F)2=G^+P-C. 

.-.  CP'=  G'+  F—  C,  and,  by  [1],  §  3, 
CP;'=(x,+  G)^+(yi+F)l 


•••  PiP  =\/{>^i  +  Gf  +  (y,-^  Ff  -  {G'  +  F' -  C), 
or  PiP-Vxr  +  yi^  +  2  Gxi  +  2  Fyi  +  C. 

Therefore,  to  find  the  length  of  the  tangent  draivn  from  a 
given  2>oint  to  a  circle,  we  have  only  to  tvrite  the  equation  of  tJie 
circle  in  the  form  x-  +  y^  +  2  Gx  +  2  Fy  +  C  =  0,  and  take  the 
square  root  of  the  result  of  substituting  the  coordinates  of  the 
23oi7it  in  the  left-hand  meynher  of  the  equation. 

Ex.  Find  the  length  of  the  tangent  drawn  from  the  point  (7,  2)  to 
the  circle  x2  +  y2  —  4x  +  6y  +  8  =  0. 


Length  =  Vt^  +  2-  -  4  (7)  +  G  (2)  +  8  =  V45  =  3  V5. 

Note.  If  the  length  of  the  tangent  as  found  by  the  above  rule  is 
zero,  the  point  is  on  the  circle ;  and  if  the  result  is  imaginary,  it  is  evident 
that  the  point  is  within  the  circle,  and  that,  hence,  no  tangent  can  be 
drawn  from  it  to  the  circle. 


58.     Chord  of  Contact. 

Frotn  any  'point  outside  a  circle  two  real  tangents  rnay  he 
draivn  to  the  circle,  and  the  straight  line  joining  their  i'esjjec- 
tive  poiiits  of  contact  is  called  the  chord  of  contact. 

Let  PiPo  and  P^Pg  be  the  tangents  to  the  circle 
x2  +  y2  +  2Gx  +  2  Fy  +  C  =  0 
from  the  point  Pi(xi,  yi).     Then  P2P3  is  the  chord  of  contact. 

Assume  the  coordinates  of  P2  and  Pg  to  be  respectively 
(x2,  ya)  and  (x^,  ys). 


THE    CIRCLE. 


109 


Fig.  50. 


By  [26],  §54,  tlie  equations  of  PjPa  and   P1P3  are  resi)ec- 
tively 

x.x  +  y.y  +  G  (x  +  X,)  +  F  (y  +  y,)  +  C  =  0,  (1) 

X3X  +  y^y  +  G  (x  +  X3)  +  F  (y  +  y3)  +  C  -  0.  (2) 

Since  Pi  is  a  point  of  both  these  lines, 

x.xj  +  y,yi  +  G  (xi  +  X,)  +  F  (y^  +  y.)  +  C  =-  0,  (3) 

X3X1  +  ysyi  +  G  (x,  +  xa)  +  F  (yi  +  y,)  +  C  =-  0.  (4) 

From  inspection  of  (3)  and  (4)  let  us  form  the  analogous 
equation 

xXi  +  yy,  +  G  (xi  +  x)  +  F(yi  +  y)  +  C  =  0.  (5) 

Of  this  equation  Ave  may  assert  two  things  : 

(1)  It  is  of  the  first  degree,  and  therefore  represents  some 
straight  line. 

(2)  It  is  satisfied  by  the  coordinates  of  both  P.,  and  Pg  by 
virtue  of  equations  (3)  and  (4). 

Therefore,  since  two  points  are    sufticient  to  determine  a 
straight  line,  (5),  or  after  re-writing. 


110  PLANE    ANALYTIC    GEOMETRY. 

XiX  +  y.y  +  G  (X  +  Xi)  +  F  (y  +  yO  +  C  =  0       [27] 

is  the  equation  of  tlie  chord  of  contact. 

As  tlie  equation  of  the  tangent  determined  by  the  point  of 
contact  and  the  equation  of  the  chord  of  contact  are  identical 
ill  form,  one  can  distinguish  between  them  only  by  noting 
whether  or  not  Pi(xi,  yi)  is  on  the  curve. 

For  example  :  Fiud  the  equation  of  the  tangent  to  the  circle 
x2  +  y2  —  4x  +  6y  +  8  =  0, 
passing  through  the  point  (1,  —  1). 

Since  (1,  —1)  satisfies  the  equation  of  the  circle,  it  is  the  point  of  con- 
tact, and  by  [26],  §  54,  the  equation  of  the  tangent  is 

Ix  -  ly  -  2  (X  +  1)  +  3  (y  -  1)  +  8  =  0, 
or  X  —  2y  —  3  =  0. 

As  a  second  example,  we  will  find  the  tangent  to  the  same  circle, 
passing  through  the  point  (7,  2). 

Since  (7,  2)  does  not  satisfy  the  equation  of  the  circle, 
7x  +  2y  -  2  (x  +  7)  +  3  (y  +  2)  +  8  =  0, 
or  X  +  y  =  0 

is  the  chord  of  contact  of  tangents  drawn  to  the  circle  from  (7,  2). 
Solving  the  equations 

x  +  y=0, 
x2  +  y2  _  4x  +  Gy  +  8  =  0, 
we  find,  as  the  points  of  contact  of  the  required  tangents,  (1,-1)  and 
(4,  -  4). 

.-.  by  [26],  §  54,  the  tangents  are 

Ix-  ly-2(x  +  1)  +  3(y  -  1)  +8  =  0, 

.  4x  -  4y  -  2  (X  -I-  4)  +  3  (y  -  4)  +  8  =  0, 

or  X  —  2y  —  3  =  0,  2x'  —  y  —  12  =  0. 

59.     Definition  and  Equation  of  the  Polar. 

It  may  seem  strange  that  the  equation  of  a  tangent  and 
the  equation  of  a  chord  of  contact  are  of  exactly  the  same 
form,  and  such  coincidence  of  form  would  be  strange  if  the 
two  classes  of  lines  were  not  more  closely  related  than  has 
been  indicated  in  the  last  articles.  As  a  matter  of  fact, 
they  are  both  but  special  cases  of  a  type  of  locus,  called  a 


THE    CIRCLE. 


Ill 


polar,  wliicli  may  be  defined  with  respect  to  a  curve  for  a 
point  which  is  called  the  pole. 

The  definition  is  as  follows  : 

If  through  a  fixed  j^oint  P^  any  number  of  secants  be  draxmi  to 
a  circle,  and  tangents  to  the  latter  be  draivn  at  the  2^0 mts  of 
intersection  of  each  secant  with  the  circle,  then  each  pair  of  tan- 
gents ivlll  intersect  upon  a  straight  line  called  the  polar  of  P^. 
Reciprocally,  Pi  is  called  the  pole  of  the  straight  line. 


Fic.  51. 

Let  the   circle   be    x-  +  y-  +  2  Gx  +  2  Fy  4-  C  =  0    and 
Pi(xi,  yO  be  the  pole. 

Let  (1)  be  any  secant  througli  P^  the  tangents  at  its  ex- 
tremities meeting  at  P2(xo,  y.,).  Then,  by  [27],  §  58,  the 
equation  of  (1)  is 

X2X  +  W  +  G  (x  +  X,)  +  F  (y  4-  y.)  +  C  =  0.  (1) 

But  Pi  is  a  point  of  line  (1)  ;  therefore 

X2X1  +  y^yi  +  G  (xi  +  X,)  +  F  (yi  +  y,,)  +  C  =  0.        (T) 

Similarly,  secant  (2)  through   Pi  determines  a  new  point 

PsCxs,  ys)  of  the  locus,  for  which  the  following  equation  is  true: 

X3X,  +  y,y,  +  G  (x,  +  X3^  +  F  (y,  +  y,.,)  +  C  =  0.       (II) 


112  PLANE    ANALYTIC    GEOMETRY. 

Comparing  equations  (I)  and  (II),  we  see  that 

xix  +  y,y  +  G  (xi  +  x)  +  F  (yi  +  y)  +  C  =  0, 

is  a  type  of  equation  satisfied  by  the  coordinates  of  every 
single  point  of  tlie  locus.     Hence 

x.x  +  yiy  +  G  (X  +  xO  +  F  (y  +  yi)  +  C  =  0       [27] 

is  the  equation  of  the  polar  of  Pi(xi,  yi)  with  respect  to  the 
CI  rclc 

xM-y'  +  2  Gx  +  2  Fy  +  C  =  0, 

(1)  If  Pi  is  on  the  circle,  it  is  evident  that  the  polar 
becomes  the  tangent  to  the  circle  at  Pj,  for  one  of  the  tan- 
gents which  meet  on  the  polar  is  always  the  tangent  at  Pj, 
so  that  all  points  of  the  polar  are  on  this  tangent  (Fig.  52). 
Hence  the  polar  and  the  tangent  coincide. 


Fig.  52. 

(2)  If  Pi  is  without  the  circle,  its  polar  becomes  the  chord 
of  contact  of  the  tangents  to  the  circle  from  Pi,  for  the  limit- 
ing positions  of  the  secants  from  Pi  are  the  two  tangents 
P1P2  and  P1P3  (Fig.  53),  and  as  the  secant  approaches  these 
tangents  as  limits,  the  corresponding  point  of  the  polar  will 


THE    CIRCLE. 


113 


approach  P2  and  P3  respectively  as  limits.  Therefore  the 
chord  of  contact  P2P3  is  the  polar  of  Pi,  since  these  two 
straight  lines  coincide  at  two  distinct  points. 


Fig.  53. 


(3)  If  Pi  is  within  the  circle,  its  polar  will  nut  cut  the 
circle  at  all  (Fig.  51),  as  is  readily  shown  by  showing  that  its 
distance  from  the  centre  of  the  circle  is  greater  than  the 
radius  of  the  circle. 


60.     Other  Properties  of  Poles  and  Polars. 

(1)  The  polar  is  perpendicnla?'  to  the  line  jobilng  the  pole  to 

X  +  G 

the  centre.     For  the  sloi)e  of  the  polar  is ^—. — -  and   the 

yi  +  r 

•    Vi  -f-  F 
slope  of  the  line  joining  the  pole  to  the  centre  is  • 

(2)  The  polar  of  the  centre  of  the  circle  is  at  an  infnite  dis- 
tance from  the  centre.  For  the  distance  of  any  point  (x,  y) 
from  the  polar  is,  by  §  32, 


114  PLANE    ANALYTIC    GEOMETIIY. 

xix  +  yiy  +  G  (X  +  xQ  -f   F  (y  +  yO  +  C^ 
V(x,+  G)^+(yi+F/  ' 

and  this  distance  becomes  infinite  if  Xi  =  —  G  and  yi  =  —  F, 
as  tliey  must  if  the  centre  is  the  pole. 

(3)    The  polar  of  any  pohit  of  a  straight  line  j^asses  through 
the  pole  of  that  line. 


Fig.  54. 

Let  Pi(xi,  yi)  be  the  pole  of  AB.     Therefore  the  equation 
of  AB  is 

xix  +  yiy  +  G  (X  +  X,)  +  F  (y  +  yO  +  C  =  0.  (l) 

If  P2(x2,  y2)  is  any  point  of  AB,  then 

X1X2  +  y^y^  +  G  (x.2  +  xO  +  F  (y^  +  y,)  +  C  =  0.     (2) 
The  polar  of  P2  will  be 

X2X  +  y.y  +  G  (x  +  X2)  +  F  (y  +  y2)  +  C  =  0.         (3) 

By  virtue   of  (2),  the   coordinates   of    Pi   satisfy   (3)  and 
hence  this  line  passes  through  P^. 

(4)  The  polar  of  the  intersection  of  two  straight  lines  is  the 
straight  line  which  joins  their  poles.      {Proof  by  (3).} 

(5)  A  triangle  ABC  is  given,  and  a  second  triangle  A'B'C  is 
draivn,  B'C  beiiig  the  p)olar  of  A,  C'A'  the  polar  of  B,  and  A'B' 


THE    CHICLE. 


115 


the  polar  of  C.  Then  A',  B',  C  are  the  respective  p)oles  of  BC, 
CA,a7idAB.     {Proof  by  (4).} 

Two  such  triangles  are  said  to  be  conjugate,  and  it  may  be 
proved  that  A  A',  BB'^  and  CC  meet  at  a  common  point. 

(6)  If  in  any  triangle  each  vertex  is  the  pole  of  the  opposite 
side,  the  triangle  and  its  conjugate  coincide,  and  it  is  said  to  be 
self-conjugate.     {Special  case  of  (5).} 

61.     Diameter. 

The  diameter  of  a  circle  is  the  locus  of  the  middle  2)oints  of  a 
system  of  parallel  chords  of  the  circle. 

Accordingly  let  us  find  the  equation  of  a  diameter  of  the 
circle 

x^  +  y^=r^  (1) 

bisecting  a  system  of  parallel  chords  of  slope  m. 


Fig.  55. 


Let  y  =  mx  +  b  (2) 

be  any  one  of  the  chords,  intersecting  tlie  circle  at  the  points 
Pi(xij  yi)  and  P2(x2,  y2)-    Then  if  K(x8j  ya)  is  the  middle  point 


116  PLANE    ANALYTIC    GEOMETRY. 

of  the  chord,  i.e.,  a  point  of  the  diameter, 

xi  +  X2  yi  +  y2  ,Qx 

^3  =  — ^  ,  y3  =  ^-y—  •  (3) 

Substituting  for  y  from  (2)  into  (1),  we  get  the  equation 
x'H"  (mx  +  b)"=  r, 
or  (1  +  m^)  x'-^  +  2m  bx  +  (b-  —  r)  =  0, 

of  which  the  roots  are  x^  and  Xo. 

Therefore  Xi  +  X2  =  —  ^Zl. — 2'  ^^  that,  by  substituting  in 
(3),  we  have 

Proceeding  in  similar  manner,  substituting  for  x  from  (2) 
into  (1),  we  find 

Equations  (4)  and  (5)  give  us  the  coordinates  of  the  middle 
point  of  any  one  of  the  system  of  chords,  according  as  differ- 
ent values  are  given  to  b.     Hence,  dividing  (5)  by  (4),  we 

get  y^  _  _  1 

X3  m' 

or  ya  = X3,  (6) 

an  equation  independent  of  b,  which  must,  therefore,  be  satis- 
fied by  the  middle  point  of  any  one  of  the  chords,  i.e.,  by  any 
point  of  the  diameter. 

Therefore  y  = x  (7) 

is  the  required  equation  of  the  diameter. 

But  (7)  is  not  a  general  equation,  since  it  is  true  only  for 
the  circle  having  its  centre  at  the  origin.  We  will  now  find 
the  general  equation  of  the  diameter  of  the  circle 

(x-d)^  +  (y-e)^  =  r2,  ^g^ 

bisecting  chords  of  slope  m. 


THE   CIRCLE. 


117 


Fig.  56. 

Transform  (8)  to  a  new  set  of  axes  parallel  to  the  old  and 
meeting  at  the  centre  of  the  circle,  Fig.  56,  by  tlie  formulas 

X  =:  d  +  x',  y  =  e  +  y'. 

The  equation  of  the  circle  becomes  x'-  +  y'-  =  r,  so  that  the 

equation  of  the  diameter  will  be,  by  (7),  y' = x',  for  the 

slope  of  the  chords  will  be  the  same  for  both  sets  of  axes, 
since  OX'  is  parallel  to  OX. 

Transforming  back  to  the  original  set  of  axes  by  formulas 
x'  =  X  —  d  and  y'  =  y  —  e,  we  have  the  equation  of  the  circle 
in  its  original  form, 

(x-d)^+(y-e)^=r^ 
while  the  corresponding  equation  of  the  diameter  becomes 


y  —  e  = (x  — d). 


[28] 


Examining  this  equation  we  find  that : 

(1)  The  diameter  is  a  straight  line. 

(2)  The  diameter  passes  through  the  centre  of  the  circle. 


118  PLANE    ANALYTIC    GEOMETRY. 

(3)  Wie  diameter  is  perpendicular  to  the  chords  which  it 
bisects. 

Any  of  these  properties  may  be  used  in  determining  a 
diameter. 

Ex.  1.     Find  the  diameter  of  the  circle  x2+ y2  +  4x  —  Gy  +  2  =  0, 
bisecting  chords  parallel  to  the  line  2x  +  y  —  1  =  0. 
The  centre  of  the  circle  is  (—2,  3)  and  m  =  —  2. 

. '.  the  diameter  is  y  —  3  =  —  ^7  (x  -|-  2)  or  x  —  2y  +  8  =  0. 

Ex.  2.  Find  the  diameter  of  the  same  circle,  which  passes  through 
the  point  (1,  —  2). 

As  before,  the  centre  of  the  circle  is  (—  2,  3). 
.*.  by  properties  (1)  and  (2),  the  diameter  is 

^^i^lOrSy  +  .«  +  >=0. 
By  property  (3),  it  bisects  chords  of  slope  f . 

62.  Circle  through  the  Points  of  Intersection  of  Two 
Circles. 

Let  x2  +  y24-2  Gx  +  2  Fy  +  C  =  0,  .      (1) 

x^  +  y^  +  2  G'x4-2  F'y  +  C'  =  0,  (2) 

be  the  two  circles. 

If  I  and  k  are  two  arbitrary  multipliers, 

l(x^  +  y^  +  2Gx  +  2Fy  +  C) 

+  k(x'^  +  y2  +  2G'x+2  F'y  +  C')  =  0  ^^ 

will  be  the  equation  of  a  circle,  since  it  is  of  the  second 
degree,  contains  no  xy  term,  and  has  the  coefficients  of  x^  and 
y^  the  same. 

Moreover,  this  circle  will  pass  through  the  points  of  inter- 
section of  circles  (1)  and  (2),  if  they  intersect.  For  if 
Pi(xi>  yO  is  a  point  of  both  circles, 

Xi^  +  yi^+2Gxi  +  2  Fyi  +  C  =  0 
and  Xi^  +  yi2  +  2  G'xi  +  2  F'yi  +  C'  =  0, 

whence  equation  (3)  will  necessarily  be  satisfied  by  (xj,  yj). 
and  therefore  circle  (3)  passes  througli  the  point  Pj. 


THE   CIRCLE.  119 

As  I  and  k  are  given  different  values,  we  shall  get  different 
circles  through  the  points  of  intersection  of  (1)  and  (2).  As 
a  special  case,  let  I  =  1  and  k  =  —  1,  when  (3)  becomes 

2  (G-G')  X +  2  (F-F')y  +  (C-C')  =  0.  (4) 

But  this  equation  is  of  the  first  degree  and  hence  is  the 
equation  of  a  straight  line.* 

In  fact,  it  is  the  equation  of  the  common  chord  of  the  two 
circles  (1)  and  (2). 

It  can  readily  be  proved  that  (4)  is  perpendicular  to  the 
line  of  centres  of  the  two  circles  (1)  and  (2). 

Note.     In  §  38  we  have  seen  that 

I  (Ax  +  By  +  C)  +  k  (A'x  +  B'y  +  C)  =  0 
is  a  straight  line  passing  through  the  point  of  intersection  of  the  two 
straight  lines 

Ax  +  By  +  C  =  0  and  A'x  +  B'y  +  C  =  0. 

Just  above  we  have  seen  that 

I  (x2  +  y2  +  2  Gx  +  2  Fy  +  C)  +  k  (x^  +  y2  +  2  G'x  +  2  F'y  +  C)  -  0 
is  a  circle  passing  through  tlie  points  of  intersection  of  the  two  circles 
x2  +  y2  +  2  Gx  +  2  Fy+  C  =  0, 
x2  +  y2  +  2  G'x  +  2  F'y  +  C  =  0. 
These  are  only  two  special  cases  of  a  more  general  theorem,  which 
may  be  stated  as  follows : 

If  U  and  V  are  any  expressions  involving  x  and  y,  U  =  0  and  V  =  0 
will  he  the  equations  of  two  curves ;  and,  if  I  and  k  are  any  two  arbitrary 
constant  multipliers,  I U  +  kV  =  0  will  he  a  curve  passing  through  all  the 
points  common  to  the  two  curves  and  intersecting  the  curves  at  no  other 
points. 

The  proof,  which  is  exactly  like  that  used  in  the  two  special  cases,  is 
left  to  the  student  to  formulate. 

63.     Eadical  Axis. 

If  the  circles  do  not  intersect,  we  can,  nevertheless,  form 
the  equation  of  the  straiglit  line, 

*  (4)  is  the  limiting  case  of  a  circle,  for  the  arc  of  a  cii-cle  approaches 
the  straight  line  as  a  limit  when  the  radius  of  the  circle  is  indefinitely 
increased. 


120 


PLANE    ANALYTIC    GEOMETRY. 


2  (G  -  G')  X  +  2  (F  -  F')  y  +  (C  -  C)  =  0, 

which  will  be  perpendicular  to  the  line  of  centres  ;  but  it  will 
not  be  the  equation  of  the  common  chord,  for  the  circles  have 
no  common  chord. 

We  shall  call  this  line  the  radical  axis  of  the  two  circles. 
A.S  obviously  this  equation  can  always  be  found,  every  pair 
of  circles  have  a  radical  axis,  which  is  their  common  chord 
if  they  intersect. 

We  will  now  prove  a  property  of  the  radical  axis,  which 
might  be  used  in  defining  it.  We  will  find  the  locus  of  points 
such  that  the  tangents  drawn  from  them  to  the  two  circles 
shall  be  of  equal  length. 


FlO.  57. 

Let  Pi(xi,  yi)  be  any  point  of  the  locus,  a  and  b  being  the 
tangents  to  the  circles  (1)  and  (2)  respectively.  If  Lj  and 
Lz  are  the  respective  lengths  of  a  and  b,  by  the  hypothesis, 

U  =  U,  or  Li^  -  U^  =  0. 

By  §57,    Lr  =  Xi^  +  yi^  +  2Gxi  +  2  Fy,  +  C, 
L/  =  Xi2  +  yi2  +  2  G'x,  +  2  F'yi  +  C, 

whence  2  (G  -  G')  Xi  +  ^  (  F  -  F')  y^  +  (C  -  C)  =  0. 


THE   CIRCLE.  121 

Therefore  Pi  is  a  point  of  the  radical  axis,  since  its  coordi- 
nates satisfy  tlie  equation  of  the  radical  axis. 

Conversely,  the  tangents  drawn  to  the  two  circles  from  any 
point  of  the  radical  axis  may  be  proved  of  equal  length,  so 
that  we  may  now  define  the  radical  axis  of  two  circles  as  the 
locus  of  2)oints  such  that  the  tangents  from  them  to  the  two 
circles  are  of  equal  length. 

64.     Radical  Centre. 

If  we  have  given  the  equations  of  three  circles, 

x^+y2  +  2  Gx  +  2  Fy  +  C  =  0,  I. 

x"  +  y-  +  2  G'x  +  2  F'y  +  C'  =  0,  XL 

x^  +  y2  +  2  G"x  +  2  F"y  +  C"  =  0,  III. 

these  three  circles  taken  in  pairs  determine  three  radical  axes, 

2  (G  -  G')  X  +  2  (F  -  F')  y  +  (C  -  C)  =  0,  (1) 

2  (G'  -  G")  X  +  2  (F'  -  F")  y  +  (C  -  C")  =  0,  (2) 

2  (G"  -  G)  X  +  2  (F"  -  F)  y  +  (C"  -  C)  =  0.  (3) 

1(2  (G  -  G')  X  +  2  (F  -  F')  y  +  (C  -  C)  }  + 

k{2  (G'  -  G")  X  +  2  (F'  -  F")  y  +  (C  -  C")}  =  0     (4) 

represents  a  straight  line  passing  through  the  point  of  inter- 
section of  radical  axes  (1)  and  (2).  In  (4)  let  I  =  —  1  and 
k  ^  —  1,  and  it  becomes 

2  (G"  -  G)  X  +  2  (F"  -  F)  y  -f  (C"  -  C)  =  0. 

But  this  is  (3);  therefore  radical  axis  (3)  passes  tlirough 
the  point  of  intersection  of  radical  axes  (1)  and  (2). 

This  common  point  of  the  three  radical  axes  is  called  the 
radical  centre  of  the  three  circles. 

It  is  evident  that,  in  general,  from  the  radical  centre  six 
equal  tangents  can  be  drawn,  two  to  each  circle. 

Ex.     Find  the  radical  centre  of  the  circles 

x2  +  y2  +  2x  —  2y  +  1  =  0,   x2  +  y2  -  8x  +  7y  —  4  =  0, 
and  X-  +  y-  +  5x  —  .'!y  +  2  =  0. 


122  PLANE    ANALYTIC    GEOMETRY. 

The    radical    axes    are    lOx  —  9y  +  5  =  0,    13x  —  lOy  +  0  =  0,    and 

3x  —  y  +  1  =  0.     The  radical  centre  is  ( — -,   —  j  and  the  common 

length  of  the  tangents  is 

65.     Polar  Equation  of  the  Circle. 

P 


Fig.  58. 

Let  C(ri,  ^i)  be  the  centre  of  the  circle  of  radius  a  and 
P(r,  0)  be  any  point  of  the  circle.     Then  Z  COP  =  ^  —  ^i. 

By  Trigonometry,  OP'  +  OC'  —  2  OP.OC  cos  ^  =  CP' 

or  r-  +  r/^  —  2rri  cos  (9  —  Gi)  =  a".  [29] 

For  any  particular  value  of  9,  the  resulting  equation  in  r  is 
of  the  second  degree,  so  that  a  line  drawn  from  the  origin 
will  meet  the  circle  in  two  distinct  points,  two  coincident 
points,  or  not  at  all,  according  as 

4  ri^  cos^  {6  —  e,)—'i  (r-"  —  a^)  is  >,  =  ,  or  <  0. 
If  4  r,^  cos^  {9  -6,)-^  (r,^  -  a")  =  0, 

ri2[l-cos2  (e-e,)^  =  a% 

sin  (0  —  e{)  =  ±  -  and  $  =  $^±1  sin~^  -. 


THE    CIRCLE.  123 


Therefore  6  must  be  as  great  as  Oi  —  sin  ^  - ,  and  as  small  as 

^1  +  sin~^  -,  if  the  corresponding  line  from  the  origin  is  to 
meet  the  circle  at  all. 

(1)  Find  the  equation  of  tlie  circle  under  the  following  conditions : 
centre  on  the  initial  line ;  centre  on  the  initial  line  and  the  circle  passmg 
through  the  origin ;  centre  on  a  luie  at  right  angles  to  the  initial  line, 
and  the  circle  passing  through  the  origin. 

(2)  What  will  be  true  of  all  the  circles  having  the  same  parameter  ri, 
but  different  parameters  a  and  Ox  ? 

(3)  What  will  be  true  of  all  the  circles  the  equations  of  which  agree 
in  ^1,  but  differ  in  ri  and  a  ? 


EXAMPLES. 
Find  the  equations  of  the  following  circles : 

1.  Centre  at  (3,  —  4)  and  radius  5. 

2.  Centre  at  (1,  3)  and  radius  2. 

3.  Centre  at  (—  2,  7)  and  radius  4. 

/      2        4\ 

4.  Centre  at  (  —  - ,  —  77  j  and  radius  6. 

5.  What  is  the  equation  of  the  circle  constructed  on  the 
line  joining  (1,  5)  and  (—3,  7)  as  diameter? 

6.  Find  the  equations  of  the  circles  of  which  the  line  join- 
ing the  points  (2,  —  3)  and  (3,  —  1)  is  a  radius. 

7.  Find  the  equation  of  the  circle  having  the  line  joining 
(—  4,  3)  and  (1,  —  2)  as  a  diameter. 

8.  Find  the  equation  of  the  circle  having  the  line  joining 
(a,  —  b)  and  (—  a,  b)  as  a  diameter. 

9.  Find  the  equation  of  the  circle  having  as  a  diameter 
that  part  of  the  line  x  — 2y  +  4==0  which  is  included  be- 
tween the  coordinate  axes. 

10.  Find  the  equations  of  the  circles  of  radius  a  which  are 
tangent  to  the  axis  of  y  at  the  origin. 


124  PLANE    ANALYTIC    GEOMETRY. 

11.  Find  the  equations  of  the  circles  of  radius  a  which  are 
tangent  to  both  coordinate  axes. 

12.  Find  the  equation  of  the  circle  having  its  centre  at  the 
origin  and  tangent  to  the  line  3x  +  4y  +  7  =  0. 

13.  Find  the  equation  of  the  circle  having  its  centre  at  the 
point  ( —  1,  2)  and  tangent  to  the  line  2x  —  y  +  3  =  0. 

14.  Find  the  equation  of  the  circle  having  its  centre  at 
(3,  —  2)  and  tangent  to  the  line  x  +  5y  +  1  =  0. 

15.  The  centre  of  a  circle  which  is  tangent  to  the  axes  of  x 
and  y  is  on  the  line  2x  —  y  +  2  =  0.     What  is  its  equation  ? 

Find  the  centre  and  the  radius  of  each  of  the  following 
circles  : 

16.  x2  +  y^-6x-2y-l  =  0. 

17.  x2  +  y-  +  4x  — 10y  +  25  =  0. 

18.  x^  +  y2-4x  +  6y  +  l=0. 

19.  3x2  +  3y2  — 6x  +  9y-2  =  0. 

20.  36x2  ^  3gy2  _  losx  +  48y  —  155  =  0. 

21.  5x2  _|_  r,y2  _  2x  +  4y  +  1  =:  0. 

22.  Ax^  4-  Ay-  +  2  Gx  +  2  Fy  +  C  =  0. 

23.  Find  the  equation  of  the  straight  line  joining  the 
centres  of  the  circles  x^  +  y^  —  4x  —  6y  —  12  =  0  and 
x'  +  y-  +  X  =  0. 

24.  Do  the  circles  4x2  +  \f  +  4x  —  12y  +  1=0, 

2x2  +  2y2  +  y  =  0 
intersect  ?     Prove  your  answer. 

Find  the  equations  of  the  circles  determined  by  the  follow- 
ing sets  of  points  : 

25.  (0,0),  (1,-1),  (-2,  3). 

26.  (2,-2),  (1,2),  (-1,1) 

27.  (0,2),  (2,0),  (-2,-2) 

28.  (a,  b),  (b,  a),  (-  b,  -  a) 

29.  (-1,2),  (3,-1),  (4,  .3) 


THE    CIRCLE.  125 

30.  Find  the  equation  of  the  circle  circumscribing  the 
triangle  (0,  3),  (—  1,  0),  (0,  -3). 

31.  Find  tlie  equation  of  a  circle  passing  through  the  point 
( —  2,  3)  and  concentric  with  the  circle 

x2+y2-3x  +  4y-2=i0. 

32.  The  centre  of  a  circle  which  passes  through  the  points 
(2,  —  1)  and  (—  1,  3)  is  on  the  line  8x  —  4y  +  5  =  0.  What 
is  the  equation  of  the  circle  ? 

33.  The  centre  of  a  circle  which  passes  through  the  points 
(2,  —  3)  and  (—  4,  —  1)  is  on  the  line  3y  +  x  —  18  =  0.  What 
is  the  equation  of  the  circle  ? 

34.  A  circle  of  radius  V85  passes  through  the  points  (2,  1) 
and  (—  3,  4).     What  is  its  equation  ? 

35.  A  circle  of  radius  Vl3  passes  through  the  points 
(2,  —  3)  and  (1,  4).     What  is  its  equation  ? 

36.  Find  the  equation  of  the  locus  of  the  centres  of  the 
circles  of  radius  3,  all  of  which  pass  through  the  point  (—1,  7). 

37.  Find  the  equation  of  the  locus  of  the  centres  of  the 
circles  which  pass  through  (a,  b)  and  ( —  b,  —  a). 

38.  Find  the  equation  of  the  locus  of  the  centres  of  all 
circles  which  pass  through  (a,  b)  and  (a',  b'). 

39.  Find  the  equation  of  a  circle  in  oblique  coordinates. 

40.  The  centre  of  a  circle  which  is  tangent  to  the  lines 
X  —  5  =  0  and  y  — 10  =  0  is  on  the  line  2x  +  3y  —  5  =  0. 
What  is  its  equation  ? 

41.  The  centre  of  a  circle  which  is  tangent  to  the  lines 

17x  +  y  —  35  =  0  and  13x  +  lly  +  50  =  0 
is  on  the  line  88 x  +  'J'Oy  +  15  =  0.     What  is  its  equation  ? 

42.  Find  the  equation  of  the  circle  circumscribing  the  tri- 
angle of  which  the  sides  are  the  lines-> 

y  +  2  =  0,  3x  -  4y  +  1  =  0,  3x  +  2y  -  5  =  0. 

43.  Find  the  equation  of  the  circle  inscribed  in  the  triangle 
the  sides  of  which  are  the  lines 

8x-f  y-17  =  0,  7x-4y  +  29  =  0,   x  +  8y  +  17=0. 


126  PLANE    ANALYTIC    GEOMETKY. 

44.  Find  the  equation  of  the  circle  inscribed  in  the  triangle 
(16,  3),  (-14,  3),  (l,-^^^ 


4 

45.  Find  a  tangent  to  the  circle  x^  +  y^  =  12,  making  an 
angle  of  30°  with  the  axis  of  x. 

46.  Find  a  tangent  to  the  circle  x^  +  y'-^  -)-  4x  ~  6y  +  9  =  0, 
making  an  angle  of  60°  with  the  axis  of  x. 

47.  Find  a  tangent  to  the  circle 

4x2  +  4y2  _  4x  +  8y  —  139  =  0, 
making  an  angle  tan~^  |  with  the  axis  of  x. 

48.  Find  a  tangent  to  the  circle 

4x2  +  4y2  _  24x  -  12y  +  25  =  0, 
parallel  to  the  line  x  —  2y  +  8  =  0. 

49.  Find  a  tangent  to  the  circle  x^  -}-  y^  —  2x  +  4y  —  4  =:  0, 
perpendicular  to  the  line  2x  —  3y  +  6  =  0. 

50.  Find  a  tangent  to  the  circle 

9x2  ^  9y2  _|.  24x  +  I8y  =  119, 
perpendicular  to  the  line  4x  +  3y  —  17  ^=  0. 

51.  What  will  be  the  slope  of  the  tangent  to  the  circle 
x2  _}_  y2  _  4x  -|_  2y  +  1  =  0,  which  passes  through  the  point 
(-2,1)? 

52.  Find  a  tangent  to  the  circle  (x  —  d)^  +  (y  —  e)^  =  r^, 
making  an  angle  tan~^  m  with  the  axis  of  x. 

53.  Find  a  tangent  to  the  circle 

x'-\-f  +  2  Gx+2  Fy  +  C  =  0, 
making  angle  tan~^  m  with  the  axis  of  x. 

54.  A  tangent  and  a  normal  to  the  circle  x^  -|-  y^  =  13  pass 
through  the  point  (2,  —  3).     What  are  their  equations  ? 

55.  A  tangent  and  a  normal  to  the  circle 

x2  +  y2  — 4x  +  2y  +  l=0 
pass  through  (2,  —  3).     What  are  their  equations  ? 

56.  Find  a  tangent  to  the  circle  x^  -f-  y^  —  8x  —  6y  —  75  =  0, 
which  passes  though  (—  10,  5). 

57.  Find  a  tangent  to  the  circle  Sx^  +  3y2  —  2x  +  4y  =  0, 
which  passes  through  (2,  1). 


THE    CIRCLE.  127 

58.  Find  a  tangent  to  the  circle  2x'  +  2y2  -f  Gx  —  6y  —  9  =  0, 

1-1               fi          1    /^l   3  +  2  V5^ 
which  passes  through  I  ~i — 

59.  Find  the  chord  of  contact  of  the  tangents  drawn  from 
(—  1,  2)  to  the  circle  x-  +  y^  —  3y  +  2  =  0. 

60.  What  is  the  length  of  the  tangent  drawn  from  (5,  4)  to 
the  circle  x^  +  y^  —  2x  —  4y  +  2  =  0  ? 

61.  What   are   the   lengths    of   the   tangents  drawn  from 


(^'2)'''''^  (2'"^) 


to  the  circle  4x- +  4y- —  20x  +  12y  +  18  =  0?     Explain  the 
second  answer. 

62.  Find  the  polar  of  the  origin  with  respect  to  the  circle 
5x2  _^  5y2  _  lox  +  7y  -  2  =r  0. 

63.  Can  real  tangents  be  drawn  from  (2,  3)  to  the  circle 

9x2  +  9y' +  6x  -  12y  +  4  =  0  ? 
Show  why  your  answer  is  correct. 

64.  Prove  that,  if  the  equation  of  the  circle  be  written  in 
the  form  (x  —  d)^+  (y  —  e)^=  r^,  the  tangent  at  Pi(xi,  yi)  is 
(xi-d)(x-d)  +  (yi-e)(y-e)  =  r2. 

65.  Does  the  polar  of  (1,  —  2)  with  respect  to  the  circle 
3x2  _^  3y2  _|_  4x  _  3y  _  -^r  _  Q  iiitersect  the  circle  ? 

Q>^.  The  polar  of  a  certain  point  with  respect  to  the  circle 
2x2  +  2y-  +  4y  +  7  =  0  is  3x  +  y  +  3  =  0.  What  are  the  co- 
ordinates of  the  point  ? 

67.  The  length  of  the  tangent  from  the  point  (x,  y)  to  the 
circle  x-'  -f  y^  -["  y  —  1  =  0  is  twice  the  length  of  the  tangent 
from  the  same  point  to  the  circle  x^  +  y^  +  3x  —  2y  +  2  =  0. 
Show  that  the  point  lies  on  the  circle 

x-  +  y-  +  4x  — ;;y-l-3  =  0. 

68.  Find  the  angle  at  which  the  circles 

x'  +  y'  +  6x  —  2y  +  5  =  0  and  x^  +  y^  +  4x  +  2y  -  5  ==  0 
intersect.     (The  angle  between  two  curves  is  the  angle  be- 
tween their  respective  tangents  at  the  point  of  intersection.) 


128  PLANE   ANALYTIC    GEOMETRY. 

69.  Show  that  the  circles  x^  +  y^  +  4x  —  6y  -|-  8  =  0  and 
x^  +  y^  —  2x  —  14y  -|-  30  =  0  are  orthogonal,  i.e.,  cut  each 
other  at  right  angles. 

70.  Find  the  length  of  the  chord  cut  from  the  line 

2x-y  +  10  =  0 
by  the  circle  x^  +  y^  +  4x  —  6y  +  11  =  0. 

71.  The  middle  point  of  a  chord  of  the  circle  x^  +  y^  =^  49  is 
(1,  3).     Find  the  equation  and  the  length  of  the  chord. 

72.  Find  the  diameter  of  the  circle 

x2  +  y2  —  2x  +  4y  +  1  =  0, 
bisecting  chords  parallel  to  y  —  2x  +  3  =  0. 

73.  Find  the  diameter  of  the  circle 

6x-^  +  6y2  +  2x  —  3y  —  5  =  0, 
bisecting  chords  which  make  an  angle  of  60°  with  the  axis 
of  x. 

74.  Find  the  diameter  of  the  circle 

x2  +  y2  4-  2x  —  lOy  +  10  =  0, 
making  an  angle  of  45°  with  the  axis  of  x. 

75.  Find  the  diameter  of  the  circle  x^  +  y"  —  x  -|-y  =  0, 
bisecting  chords  perpendicular  to  3x  4"  2y  +  1  =  0. 

76.  A  diameter  of  the  circle 

4x'-'  +  4y-  +  8x  -  12y  +  1  =  0 
passes  through  the  point  (1,  —  1).     What  is  its  equation  and 
the  slope  of  the  chords  which  it  bisects  ? 

77.  Find  the  radical  axis  of  the  two  circles 

2x2  _|_  2y2  +  3x  —  4y  +  1  =  0  and  3x=^  +  3y2  -  6x  +  1  =  0. 

78.  Find  the  radical  axis  and  the  line  of  centres   of  the 

two  circles 

x2  +  y2  — 4x  +  6y-2  =  0 
and  2x^  +  2y2+6x-8y  +  l  =  0, 

and  show  that  they  are  perpendicular  to  each  other. 

79.  Find  the  point  of  intersection  of  the  radical  axis  and 
the  line  of  centres  of  the  two  circles 

x^  -|-  y^  =  25    and    x-  +  y-  —  6x  —  8y  =  24. 


THE   CIRCLE.  129 

80.  Find  the  radical  centre  of  the  circles 

x'  +  y'  +  6x  +  y  —  5  =  0, 
2x2  +  2y2-x-3y  +  l  =  0, 
x"  +  y2+7x  +  4y-2  =  0. 

81.  Find  the  circle  which  passes  through  (0,  4)  and  the 
points  of  intersection  of  the  two  circles  x^  +  y'  =  10  and 
x2  +  y2-8x  +  2y  +  8  =  0. 

82.  Show  that  the  circles  x2  +  y2  +  2  Gx  +  2  Fy  +  C  =  0 
and  x:'  +  y2  +  2  G'x  +  2  F'y +  C'  =  0  are  tangent  to  each 
other,  if 


V(G  -  G')^  +  (F  -  F')-  =  VG-  +  F^  -  C  ±  VG'^  +  F'-  -  C. 

83.  Prove  that  the  radical  axis  of  any  two  circles  is  per- 
pendicular to  the  line  joining  their  centres. 

84.  Prove  that  two  circles  will  be  concentric  if  their  equa- 
tions differ  only  in  the  absolute  term. 

85.  If  A,  B,  C,  and  D  are  four  points  on  a  circle,  and  the 
tangents  at  A  and  B  meet  at  P,  and  the  tangents  at  C  and  D 
meet  at  Q,  and  the  chords  A  B  and  C  D  meet  at  R,  prove  that  R 
is  the  pole  of  PQ. 

86.  The  polars  of  A(xi,  yi)  and  B(x2,  y2)  are  given  with 
respect  to  the  circle  x^  +  y^  =  rl  A  P  is  the  perpendicular 
from  A  to  the  polar  of  B,  and  BQ  is  the  perpendicular  from 
B  to  the  polar  of  A.     If  C  is  the  centre  of  the  circle,  prove 

fi    f  CA      CB 

^^^'"'  AP=BQ- 

87.  Lines  are  drawn  from  ( —  a,  0)  and  (a,  0)  at  right 
angles  to  each  other.  Find  the  equation  of  the  locus  of  their 
point  of  intersection. 

88.  Prove  that  the  square  of  the  tangent  drawn  from  any 
point  of  one  circle  to  another  circle  is  proportional  to  the 
perpendicular  from  that  point  to  their  radical  axis. 

89.  Find  the  locus  of  a  point  such  that  tangents  from  it  to 
two  concentric  circles  are  inversely  as  their  radii. 


130  PLANE   ANALYTIC    GEOMETRY. 

90.  A  point  moves  so  that  the  sum  of  the  squares  of  its 
distances  from  any  number  of  fixed  points  is  constant.  Show 
that  the  locus  is  a  circle. 

91.  A  point  moves  so  that  its  distances  from  two  fixed 
points  are  in  a  constant  ratio  k.  Show  that  the  locus  is  a 
circle,  except  when  k  =  1. 

92.  A  line  of  length  a  moves  with  its  extremities  upon  the 
coordinate  axes.  Find  the  locus  described  by  its  middle 
point. 

93.  0  is  a  fixed  point  and  AB  is  a  fixed  straight  line.  A 
straight  line  is  drawn  from  0  meeting  AB  at  P,  and  in  OP  a 
point  Q  is  taken  so  that  OP.  OQ  =^  k'.     Find  the  locus  of  Q. 

94.  A  point  moves  so  that  the  length  of  the  tangent  from 
it  to  a  fixed  circle  is  always  equal  to  its  distance  from  a  fixed 
point.     Find  the  locus. 

95.  A  point  moves  so  that  the  sum  of  the  squares  of  its 
distances  from  the  four  sides  of  a  square  is  constant.  Show 
that  the  locus  is  a  circle. 

96.  A  point  moves  so  that  the  square  of  its  distance  from 
the  base  of  an  isosceles  triangle  is  equal  to  the  product  of  its 
distances  from  the  other  two  sides.  Show  that  the  locus  is 
a  circle. 

97.  A  point  moves  so  that  the  sum  of  the  squares  of  its 
distances  from  the  sides  of  an  equilateral  triangle  is  constant. 
Show  that  the  locus  is  a  circle. 

98.  Find  the  locus  of  a  point,  the  square  of  the  distance  of 
which  from  a  fixed  point  is  proportional  to  its  distance  from 
a  fixed  straight  line. 


CHAPTER    VI. 

THE    CONIC    SECTIONS. 

66.     Definition  and  Equation. 

A  conic  section  is  the  locus  of  a  point  ivhich  moves  so  that 
its  distance  from  a  fixed  point,  called  the  focus,  is  in  a 
constant  ratio  to  its  distance  from  a  fixed  straight  line,  called 
the  directrix. 

The  constant  ratio  is  called  the  eccentricity  of  the  conic 
section,  and  will  be  denoted  thronghout  this  book  by  the  letter 
e.  Its  value  is  of  great  importance,  not  only  in  lixing  the 
shape  of  any  one  conic  section,  but  also  in  determining  to 
which  one  of  three  kinds  the  curve  belongs.  Jfe<Cl,  the 
co)iic  section  is  called  an  ellipse;  if  e  =  1,  it  is  called  a 
parabola  ;  if  e  !>  1,  it  is  called  an  hyperbola. 

As  the  name  literally  indicates,  the  conic  sections  may  be 
obtained  by  cutting  a  right  circular  cone  by  a  plane,  and  it 
was  from  this  standpoint  that  the  curves  were  first  studied 
and  named.  For  the  purposes  of  Analytic  Geometry,  how- 
ever, it  is  more  convenient  to  use  the  definition  given 
above.  We  Avill  later  prove  that  the  curves  thus  defined  are 
identical  with  those  obtained  from  the  cone.  In  anticipation 
of  this  proof,  we  will  continue  the  use  of  the  name  "  conic 
section,"  or  more  shortly,  "  conic." 

We  can  readily  deduce  the  equation  of  the  conic  section 
from  the  definition. 

Let  RS  (Fig.  59)  be  tlie  directrix  and  F  the  focus.  Take 
RS  as  the  axis  of  y,  and  a  line  through  F  perpendicular  to 
RS  as  the  axis  of  x.  Then  D,  the  intersection  of  these  two 
lines,  is  the  origin  of  coordinates.  Call  the  distance  DF,  2p. 
Then  the  coordinates  of  F  are  (2p,  0). 


132 


PLANE   ANALYTIC    GEOMETRY. 


c 

5 

M 

P 

/ 

• 

D 

f/ 

N 

Fig.  59. 


Let  P  be  any  point  on  the  conic  section,  DN  its  abscissa,  x, 
and  NP  its  ordinate,  y.  Draw  FP,  and  PM  perpendicular  to 
RS.     Then,  from  the  definition, 

PF_ 
PM~^' 

or  PF'  =  e^  PM'. 

But  PM  =  X,  the  ordinate  of  P,  and  by  [1],  §  3, 

PF'=(x-2p)2+y2. 
Therefore 

(x-2p)^  +  y'  =  eV  [30] 

is  the  required  equation  of  the  conic  section. 

If  we  place  x  =  0  in  equation  [30],  we  find  the  corre- 
sponding value  of  y  to  be  imaginary.  Hence,  a  conic  section 
never  i7iter sects  the  directrix. 

The  farther  discussion  of  the  conic  sections  is  best  made 
by  considering  each  case  separately. 


THE   CONIC   SECTIONS. 


133 


67.     The  Parabola,    [e  =  1] 
S 


Fig.  60. 

If  we  place  e  =  1  in  equation  [30],  it  becomes 
(x-2p)2  +  y■^=x^ 
or  y"  —  4px  +  4p-^0. 

By  placing  y  =  0  in  this  equation,  we  find  that  the  curve 
intersects  the  axis  of  x  in  one  point,  namely  x  =  p.  This 
point  is  called  the  vertex  (A)  of  the  parabola,  and  the  line 
(AF)  drawn  through  the  focus  and  the  vertex  is  called  the  axis. 
Remembering  that  the  distance  from  the  directrix  to  the  focus 
is  by  definition  2p,  we  see  that  the  jxirahola  /las  one  vertex 
half  way  between  the  directrix  and  the  focus. 

The  equation  of  the  parabola  will  take  a  simpler  form,  if 
the  origin  of  coordinates  be  transferred  to  the  vertex,  the 
axis  of  X  being  unchanged,  but  a  new  axis  of  y  being  taken 
through  A.  This  transformation  is  effected  by  the  substitu- 
tion [19],  §  43, 


134  PLANE  ANALYTIC    GEOMETRY. 

x  =  x'  +  p, 

y  =  y'. 
There  results 

y'2  =  4px', 
or,  dropping  the  primes, 

y'  =  4px.  [31] 

This  is  the  simplest  equation  of  the  parabola. 

From  this  equation,  the  parabola  may  be  readily  plotted  by 
computing  ordinates  corresponding  to  values  of  x.  Among 
these  ordinates,  the  one  passing  through  the  focus  is  to  be 
especially  noticed. 

The  abscissa  of  the  focus  is  p,  tlie  corresponding  ordinate 
is  ±2p.  Hence  the  line  LL'  =  4p.  This  line,  the  chord 
through  the  focus  perpendicular  to  the  axis,  is  called  the 
latus  rectum  of  the  curve. 

If  p  is  positive,  the  equation 

y^  =  4px 

represents  a  curve  as  in  Fig.  61  (a).  If  p  is  negative,  the 
same  equation  represents  a  curve  as  in  Fig.  61  (b). 

If  we  had  called  the  directrix  the  axis  of  x  at  the  outset, 
the  equation  of  the  parabola  would  have  reduced  to  the  form 
x2  =  4py.  In  this  case,  the  curve  would  be  in  the  position 
represented  in  Figs.  61  (c)  and  61  (d),  respectively,  according 
as  p  is  positive  or  negative. 

In  connection  with  any  parabola,  we  should  notice  its  axis, 
directrix,  and  focus.  If  the  equation  is  in  the  form  y^  =  4px, 
the  equation  of  the  axis  is  y  —  0  ;  the  equation  of  the  direc- 
trix is  X  =  —  p  ;  and  the  focus  as  at  the  point  (p,  0). 

If  the  equation  is  in  the  form  x^  =  4py,  the  equation  of 
the  axis  is  x  =  0  ;  the  equation  of  the  directrix  is  y  ^  —  p  ; 
and  the  focus  is  at  the  point  (0,  p). 

Ex.  Find  the  equations  of  axis  and  directrix,  and  the  focus  of  the 
parabola  y^  =  —  7x. 


THE    CONIC    SECTIONS. 


135 


Here 


.:  Focus  is  at  (—  |,  0) ; 
Equation  of  axis,  y  =  0 ; 


Fig.  61. 


(b) 


(d) 


68.     The  Ellipse,    [e  <  1] 

In  interpreting  equation  [30],  we  have  to  remember  tliat 
e  is  less  than  unity.  Placing  y  i=  0,  we  have  two  values  of 
X,  corresponding  to  two  vertices  A  and  A',  Fig.  62,  namely, 


2p 


2p 
1-e' 


-^^      1  +  e' 

Since  e  <  1,  Xi  <  2p  and  Xo  >  2p.  Hence,  the  ellipse  has 
two  vertices  lyiiKj  on  the  same  side  of  the  directrix  but  on  oiqyo- 
site  sides  of  the  foctis. 

The  line  A  A'  is  called  the  major  or  transverse  axis  of  the 

ellipse,  and  its  length  is  denoted  by  2a. 


136 


PLANE   ANALYTIC   GEOMETRY. 


Fig.  62. 


Then 


whence 


2a  =  DA' -DA, 
_    2p  2p 


4pe 

1  — e      l  +  e~l-e2' 
_    2pe_ 


(1) 


The  point  C,  half-way  between  A  and  A',  is  called  the 
centre  of  the  conic.  Its  distances,  both  from  the  directrix 
and  from  the  focus,  are  readily  found. 


For 


DC 


DA  +  AC, 


l  +  e 
_     2p 


,     2pe    ^     2p 
l  +  e^l-e^      l-e^' 


Hence 
Also 

Hence 


DC  = 


FC=DC 

9 


8 

DF, 


(2) 


-^^_2p-lP^ 


l-e^ 
FC=  ae. 


1-6^ 


(3) 


THE   CONIC    SECTIONS.  137 

The  equation  of  the  ellipse  will  take  its  simplest  form  if 
the  origin  of  coordinates  is  transformed  to  C,  the  direction 
of  the  axis  being  unchanged.  For  that  purpose,  we  substi- 
tute in  equation  [30],  the  values 

e  .«. 


and  also,  from  (1), 
There  results 


y  =  y 


Zp^= ae. 

e 


(x' +  ae)'^  +  y"^  =  eM  x'  + 


e 


Whence,  by  expanding,  uniting  like  terms,  and  dropping 
the  primes, 

(l-e^)x^'  +  y^-a'^(l-e^), 

a"^+aMl-e^)^^- 
If,  now,  we  introduce  a  new  quantity  b,  such  that 

b2  =  a2  (1-e'^),  (4) 

we  have,  finally, 

$  +  $=h  [32] 

the  simplest  equation  of  the  ellipse. 

From  this  equation  the  ellipse  may  be  plotted,  and  the 
shape  given  in  tlie  figure  may  be  verified.  The  meaning  of 
the  quantity  b  is  found  by  placing  x  =:  0  in  [32].  There 
results  y  =  ±  b,  and  hence  the  line  BB'  is  equal  to  2b.  It  is 
evident  from  (4)  that  b  <  a,  since  e  <  1  ;  hence  BB'  is  called 
the  minor  axis,  or  sometimes  the  conjugate  axis. 

When  the  semi-axes  a  and  b  are  given,  the  eccentricity  is 
readily  computed.     For,  from  (4),  we  have 

Va^  -  b'^  ... 

6= .  (O) 


138  PLANE   ANALYTIC    GEOMETRY. 

From  [32]  it  is  evident  that  the  curve  is  symmetrical  with 
respect  to  the  axis  of  y.  Hence,  if  the  left-hand  portion  of 
Fig.  62  is  folded  over  upon  tlie  axis  of  y,  it  will  come  into 
exact  coincidence  with  the  right-hand  portion  of  the  figure. 
It  follows  that  the  ellipse  has  another  focus  F'  and  another 
directrix  R'S'  symmetrical  to  F  and  RS. 

The  latus  rectum  of  an  ellipse  is  defined  as  the  chord 
through  the  focus  perpendicular  to  the  major  axis.  To  find 
its  length,  we  place  x  ==  ae  =  Va^  —  b^  in  [32],  and  find 

b'      XT  ,  ,  ,      2b-' 

y  =  ±  — .     Hence  LL  —  —  . 
a  a 

We  have  discussed  the  equation  of  the  ellipse  up  to  this 
point  on  the  assumption  that  the  directrix  RS  was  taken  as 
the  axis  of  y  in  the  deduction  of  the  original  equation  [30]. 
It  would  have  been  equally  allowable  to  take  RS  as  the  axis 
of  X,  however,  and  to  call  the  major  axis  2b,  the  minor  axis 
2a.     We  should  be  led  to  the  same  final  equation 

a-'  ^  b^ 
as  before,  but  in  all  the  intermediate  steps  the  quantities  x 
and  y,  and  a  and  b  would  interchange  their  roles.     This  result 
is  at  once  distinguished  from  [32]  from  the  fact  that  here 
b  >  a,  whereas  in  [32],  a  >  b. 

For  convenience,  we  collect  our  results  as  follows : 

I.   Corresponding  to  the  ellipse 

where  a  !>  b,  are  the  following  formulas : 
Major  axis,  y  =  0  ; 
Centre,  (0,  0)  ; 
Va^*  -  b^ . 


a 

Foci,  (±  ae,  0)  ; 


THE   CONIC   SECTIONS.  139 

Vertices,  (±  a,  0)  ; 

^.         ,  a 

Directrices,  x  =  ±  -  : 
e 


2b2 
Latus  rectum,  — . 
a 

II.   Corresponding  to  the  ellipse 


a^  '  b 


_2     I^U2  -^' 


ivhere  a  <  b,  are  the  following  formulas : 
Major  axis,  x  =  0  ; 
Centre,  (0,  0) ; 
Vb^  -  a^ 


b        ' 
Foci,  (0,  ±  be)  ; 
Vertices,  (0,  ±  b)  ; 

b 

Directrices,  y  :=±-: 
^  e 

2a2 
Latus  rectum,    7—. 
b 

If  a  =  b,  the  equation  of  the  ellipse  reduces  to 

x^  +  y^  =  a^ 

the  equation  of  a  circle.      By  substituting  b  ^  a  in  the  above 

formulas,  we  find  e  =  0  ;  foci,  (0, 0) ;  directrices,  y  =  ±  -  =  00. 

The  last  result  must  be  interpreted  by  the  method  of  limits, 
and  we  say  :  the  circle  is  the  limiting  form  uihich  the  ellipse 
approaches  as  the  eccentricity  approaches  zero  and  the  directrix 
recedes  indefinitely  from  the  origin. 

Consider,  as  example,  i)x-  +  4y'-  =  36. 

This  is  equivalent  to 

—  +  —  =  1 
whence  a  =  2,  b  =  3,  and  a  <  b. 


140 


PLANE   ANALYTIC   GEOMETRY. 


We  take  the  second  set  of  formulas  with  the  results : 
Major  axis,  x  =  0; 
Centre,  (0,  0) ; 
e=iV5; 
Foci,  (0,  ±  V5) ; 
Vertices,  (0,  ±  3) ; 
Directrices,  y  =  ±  |  Vs ; 
Latus  rectum,  f . 

69.     The  Hyperbola,     [e  >  1] 


D' 


Fig.  63. 


The  equation  [30]  is  to  be  interpreted  on  the  hypothesis 
that  e  is  greater  than  unity.  Placing  y  r=  0,  we  have  for  the 
vertices 


2p 


-P 


'      1  +  e'    ^      1-e 
Since  e  >  1,  Xi  <  2p  and  Xg  is  negative. 
Hence,  the  hyperbola  has  tivo  vertices  lying  on  the  same  side 
of  the  focus  and  on  opposite  sides  of  the  directrix. 

The  line  A'A  is  called  the  transverse  axis  of  the  hyperbola 
and   its   length  is   denoted  by   2a.     The  point    C,  half-way 
between  A'  and  A,  is  the  centre  of  the  curve. 
Then  2a  =  -DA'+DA, 

2p  2p    _    4pe 

■~      l-e^l  +  e~e^^=^' 


THE   CONIC    SECTIONS.  141 

whence  a  =   .-,  '^   -  •  (1) 

e-  —  1 

DC  =  DA  +  AC, 

_    2p  2pe    _         2p 

~l  +  e       e^-l"      e--l' 

a 

e 

FC=FD+DC, 


whence  DC  =: (2) 

e 


o,  2p     _        2pe^ 

-1~     e'-r 


whence  FC  ^  —  ae.  (3) 

The  origin  may  be  transferred  to  C  by  placing 

-      a 
e 

and  2p  =  ae 

e 

There  results,  as  in  the  case  of  the  ellipse, 

a2^a2(l-e'-^) 

The  denominator  nnder  y-  is  now  a  negative  quantity,  since 
e  >  1.     We  will,  therefore,  place 

—  h^=a'(l-e''), 
or  b''=a-(e^-l).  (4) 

The  equation  then  becomes 

r!-^=l,  [33] 

a       b 

the  simplest  equation  of  the  hyperbola. 

From  this  equation  the  hyperbola  is  easily  plotted.  It  will 
be  found  to  consist  of  two  infinite  branches,  cutting  the  axis 
of  X  at  the  points  A  and  A'. 

When  X  =  0,  y  =  ±  b  V—  1.  This  is  imaginary,  but  it  is 
convenient  to  lay  off  the  ordinates  CB  and  CB',  each  equal  to 


142 


PLANE   ANALYTIC    GEOMETRY. 


b,  and,  from  analogy  to  the  ellipse,  to  call  the  line  BB'  the 
conjugate  axis.  This  axis  appears  at  first  sight  to  have  no 
connection  with  the  hyperbola,  but  the  following  discussion 
shows  that  it  has  an  important  part  in  determining  the  shape 
of  the  curve. 

Consider  a  line,  y  =  mx,  passing  through  the  centre.     It 
cuts  the  hyperbola  in  two  points,  the  abscissas  of  which  are 

ab 
x  =  + 


Vb^- 


a-m'' 


These  values  of  x  are  real  if  m-<— ,  and  imaginary  if 

3." 


a 


m'^>  -5.     The  dividing  case  between  these  is  when  m  : 
a'' 

corresponding  to  the  two  straight  lines 

b  ,  b 

y  ^  -X         and         y  = x. 

a  ■'a 

If,  now,  we  construct  a  rectangle  of  which  the  centre  is  at 

C  and  the  sides  are  equal  to  2a  and  2b,  respectively,  these 

two  lines  will  be  its  diagonals.    Any  line  the  slope  of  which 

is  less  than  that  of  these  diagonals  will  cut  the  hyperbola 

in  two  points ;  any  line  the  slope  of  which  is  greater  than 

that  of  the  diagonals  will  not  intersect  the  hyperbola. 


Fig.  64. 


THE   CONIC    SECTIONS.  143 

Hence  these  diagonals  'tnay  he  used  as  guide  lines  in  draiving 
the  branches  of  the  hyperbola.  They  are  called  the  asymptotes 
of  the  hyperbola. 

If  the  semi-axes  a  and  b  of  the  hyperbola  are  given,  the 
eccentricity  is  given  by  the  formula 

^_  Va^+b^^  (5) 


a 
as  follows  at  once  from  (4). 

The  symmetry  of  the  hyperbola  shows  the  existence  of  a 
second  focus  and  directrix  at  the  left  of  the  centre. 

The  formulas  thus  far  obtained  depend  upon  the  original 
assumption  of  the  directrix  as  the  axis  of  y  in  the  derivation 
of  formula  [30].  If  the  directrix  had  then  been  taken  as  the 
axis  of  X,  the  simplest  equation  would  be 

a^^b^        ' 
and  in  all  the  discussion  x  and  y,  and  a  and  b  would  in- 
terchange  their   roles.     We   have,   therefore,  the    following 
summary  of  our  work. 

I.   Corres2)onding  to  the  hyperbola 

a^      b'^        ' 

are  the  folloiving  formulas : 

Transverse  axis,  y  ^  0  ; 
Centre,  (0,  0) ; 

Va'^  +  b^ 

^  = 1~' 

Foci,  {±L  ae,  0)  ; 
Vertices,  (db  a,  0) ;  ' 

Directrices,  x  =  ±  - ; 

Lotus  rectum,  — . 
a 


144  PLANE   ANALYTIC   GEOMETRY. 


II.   Corresponding  to  the  hyperbola 

are  the  following  formulas  : 

Transverse  axis,  x  =  0  ; 
Centre,  (0,  0)  ; 

Vb^  +  a^ 
e  = 


b 

Foci,  (0,  ±  be)  ; 

Vertices,  (0,  ±  b)  ; 

b 
Directrices,    y  =  ±  — ; 
e 

Asymptotes,  y  =:  ±  -x  ; 
a 

2a^ 
Latus  red  ion,  -r-  • 
b 

70.     Generalization. 

We  have  seen  that,  by  proper  choice  of  coordinate  axes,  the 
equations  of  the  three  conic  sections  may  be  reduced  to  forms 
of  the  types 

y2  =  4px, 

^  1  y!_i 

a==^b^~-^' 


r_y_- 

If  these  three  equations  are  transferred  to  new  axes  parallel 
to  the  old,  the  new  origin  being  at  the  point  (xq,  yo),  by  means 
of  [19],  §43,  they  become 

(y'  +  yo)^  =  4p(x'  +  xo), 

(x'+xo)^  ,  (y'  +  yo)^^-. 
a^         '  b'^  ' 

(x'+x„y^      (y'+yoy^^., 

a-  b^ 


THE    CONIC    SECTIONS.  145 

If  the  above  equations  are  simplified,  they  will  all  be  found 
to  be  of  the  general  form 

Ax^  +  By2  +  2  Gx  +  2  Fy  +  C  =  0. 

Conversely,  an  equation  of  the  form  just  written  can  be 
put  in  the  form  of  the  equation  of  one  of  the  conic  sections, 
except  in  certain  exceptional  cases.  The  theoretical  discus- 
sion will  be  found  in  chapter  XI ;  it  is  the  purpose  of  the 
present  chapter  to  give  methods  by  which  the  student  may 
accomplish  the  required  results  with  the  least  possible  de- 
pendence upon  formulas.  This  will  best  be  done  by  consider- 
ing a  few  numerical  examples.  It  will  be  seen  that  the 
process  is  a  modification  of  that  called  "completing  the 
square "  in  the  solution  of  quadratic  equations. 

Ex.   1.     Take  the  equation 

2x2  +  3y^  -h  8x  -  6y  +  5  =  0. 
It  may  be  written 

2(x2  -I-  4x)  -F  3  (y2  -  2y)  =  -  5. 
We  may  now  complete  the  square  of  the  quantity  in  eacli  parenthesis, 
adding  to  the  right-hand  side  of  the  equation  the  same  quantities  that 
are  added  to  the  left-hand  side. 
There  results 

2  (x'-i  -t-  4x  -f-  4)  -f  3  (y2  -  2y  4-  1)  =  -  5  -h  8  +  3, 
or  2  (X  -I-  2)2  +  3  (y  -  1)2  =  6. 

If  we  now  place 

X  -f-  2  =  x', 

y  - 1  =  y'> 

thus  transforming  the  equation  to  new  axes  parallel  to  the  old  ([19], 
§  43),  the  new  origin  being  at  the  point  (—  2,  1),  we  have 
2x'2  +  3y'2  =  6, 
x'2  ,  y'2 

the  equation  of  an  ellipse,  in  which  a  =  Vs,  b  =  \/2.     Since  a  >>  b,  we 
apply  the  first  set  of  formulas  of  8  68,  with  the  following  results.- 

Centre,  (x'  =  0,  y'  =  0) ; 
Major  axis,  y'  =  0 ; 


146 


PLANE    ANALYTIC    GEOMETRY. 


Vertices,  (x'  =  ±  V3,  y'  =  0)  ; 
Foci,  (x'  =  ±  1 ,  y'  =  0)  ; 
Directrices,  x'  =  ±  3. 
If  we  now  transfer  these  results  back  to  the  old  axes  x  and  y,  we  have, 

finally. 

Centre,  (-  2,  1)  ; 
Major  axis,  y  =  1  ; 

e  =  V|; 
Vertices,  {—  2  ±  Vs,  1)  ; 
Foci,  (-  1,  1)  and  {-  3,  1) ; 
Directrices,  x  =  + 1  and  x  =  —  5. 


Ex.  2.  Let  us  consider    2x2  _  3y2  -|-  gx  -  6y  +  11  =  0. 
As  before,  we  have  the  following  work : 


Now  place 


we  have 


2  (x2  +  4x)  -  3  (y2  +  2y)  =  -  11, 
2  (x2  +  4x  +  4)  -  3  (y2  +  2y  +  1) 
2(x  +  2)2-3(y  +  1)2  =  -6. 

X  +  2  =  x', 

y  +  1  =  /; 

2x'2  —  3y'2  =  —  6, 


11  +  8-3, 


The  curve  is  therefore  an  hyperbola,  with  the  tranverse  axis  parallel 
to  OY.     We  find  by  §  69  the  following  values : 

Centre,  (x'  =  0,  y'  =  0) ; 


THE   CONIC    SECTIONS. 


147 


Transverse  axis^  x'  =  Oj_ 
e  =  Vf  =  WlO; 
Foci,  (x'  =  0,  y'  =  ±  V5) ; 
Vertices,  (x'  =  0,  y'  =  ±  V2) ; 
Directrices,  y'  =  ±  ?  Vo ; 
Asymptotes,  x'  =  ±  Vfy'. 
These  results,  transferred  back  to  tlie  old  axes  OX  and  OY,  are 
Centre,  (-  2,  —  1) ; 
Transverse  axis,  x  =^  2 ; 

e  rr  1  VlO  ; 
Foci,  (—  2,  —  1  ±  V5) ; 
Vertices,  (-  2,  —  1  ±  V2) ; 
Directrices,  y  =  —  1  ±  |  V5 ; 
Asymptotes,  x  +  2  =  ±  Vi(y  +  1). 


Fig.  66. 

Ex.  3.    Finally,  we  will  consider 

y2  —  4y  +  8x  +  28  =  0. 
There  is  here  only  one  quadratic  term.     By  completing  the  square, 
we  have 


148 


PLANE   ANALYTIC   GEOMETRY. 


y2  —  4y  +  4  =  —  8x  —  24, 
or  (y  -  2)-2  =  -  8  (x  +  3). 

Placing 

•  y  -  2  =  y', 

X  +  3  =  x', 
we  have 

y'2  -  -  8x'. 

This  is  the  equation  of  a  parabola,  in  which  4p  =  — 
By  §  67,  w^e  have  the  results  : 

Vertex,  (x'  =  0,  y'  =  0)  ; 

Axis,  y'  =  0 ; 


Pocus,  (x' 


2,  y'-O); 


Directrix,  x'  =  2. 
Transforming  hack  to  the  original  axes,  we  have 
Vertex,  (-  3,  +  2) ; 
Axis,  y  =  2  ; 
Focus,  (—  5,  2)  ; 
Directrix,  x  =  —  1. 


Fig.  67. 


THE    CONIC    SECTIONS.  149 

71.     Limiting^  Cases  of  Conic  Sections. 

There  are  certain  cases  where  the  process  of  completing 
the  squares,  as  used  in  the  previous  article,  leads  to  a 
result  which  is  not  one  of  the  equations  of  the  ellipse,  hyper- 
bola, or  parabola,  although  bearing  a  strong  similarity  to 
one  of  them.     These  cases  are  as  follows : 

1.  Similar  to  the  Ellipse. 

On  completing  the  squares,  we  may  be  led  to  the  form 

a-^  ^  b^    . 

This  equation  is  satisfied  by  no  real  values  of  x'  and  y',  since 
the  sum  of  two  squares  cannot  be  negative.  There  is,  there- 
fore, no  real  curve  corresponding  to  the  equation  ;  but,  because 
of  the  form  of  the  equation,  it  is  sometimes  said  to  represent 
an  ^'  imafjinary  ellipse.'''' 

In  the  second  place,  on  completing  the  squares,  we  may 
find  the  right-hand  side  equal  to  zero. 

We  have,  then, 

—  4-^  =  0. 
a''       b- 

Tliis  is  satisfied,  as  far  as  real  quantities  are  concerned, 
only  by 

x'  =  0,         y'  =  0, 

for  the  sum  of  two  squares  can  be  zero  only  when  each  of  the 
squares  is  zero.    Hence  the  equation  represents  a  sinyle  point.* 

We  may  say,  therefore  : 

The  iynaginary  ellipse  and  the  point  are  exceptional  cases  of 
the  ellipse. 

2.  Similar  to  the  Hyperbola. 

On  completing  the  squares,  the  right-hand  side  of  the 
equation  may  be  zero. 

*  This  equation  may  also  be  said  to  represent  two  imaginary  straight 
lines  intersecting  in  a  point.     See  §  41  and  §  133. 


150  PLANE   ANALYTIC    GEOMETRY. 

The  equation  is  tlien  of  the  form 

0. 


X      y_:  _ 


This  is  factored  into 

n')(^-0=«. 

and  each  factor  represents  a  straight  line  passing  through  the 
point  (0,  0). 

Hence  :  Two  intersecting  strut (jltt  lines  are  an  exceptional 
case  of  an  hyperhola. 

3.   Similar  to  the  Parabola. 

An  exceptional  case  arises  here  when  the  equation  con- 
tains only  one  of  the  coordinates,  let  us  say  y. 

In  this  case  the  left-hand  side  may  be  factored  into  two 
linear  factors, 

(y--)(y-/3)  =  o. 

Each  factor  represents  a  line  parallel  to  the  axis  of  x,  and 
the  two  are  distinct,  coincident,  or  imaginary,  according  as  a 
and  /8  are  real  and  distinct,  equal,  or  imaginary. 

Hence  :  Two  parallel  straight  lines,  which  may  he  real,  coin- 
cident, or  imaginary,  are  excejitional  cases  of  the  parabola. 

72.     Recapitulation. 

The  results  of  the  foregoing  articles  lead  to  the  following 
statement : 

A  quadratic  equation  of  the  form 

Ax-  +  By-  +  2  Gx  +  2  Fy  +  C  =  0 

always  represents  a  conic  section. 

(1)  If  f^  and  B  have  the  same  sign,  the  conic  is  an  ellipse,  a 
circle,  or,  in  exceptional  cases,  an  imaginary  ellipse  or  a  point. 

(2)  If  A  and  B  have  opposite  signs,  the  conic  is  an  hyperbola, 
or,  in  exceptional  cases,  tivo  intersecting  straight  lines. 


THE   CONIC    SECTIONS.  151 

(3)  If  either  A  or  B  is  zero,  the  conic  is  a  parabola,  or,  in 
excejitional  cases,  two  ^ja>'a^^eZ  straight  lines  which  may  be  real, 
coincidejit,  or  imaginari/. 

The  student  should  notice  that  the  above  equation  is  not 
the  most  general  equation  of  the  second  degree,  since  it  lacks 
the  term  2  Hxy.  The  discussion  of  the  general  equation  is 
found  in  chapter  XI. 

73.     Sections  of  a  Cone. 

In  this  article  we  mean  to  prove  the  statement  made  in 
§  66  of  this  chapter  that  every  plane  section  of  a  right 
circular  cone  is  one  of  the  curves  we  have  called  the  conic 
sections.  In  so  doing  we  need  to  distinguish  the  case  in 
which  the  plane  that  cuts  out  the  section  passes  through 
the  vertex  from  that  in  which  it  does  not.  In  the  latter  case 
we  shall  prove  the  theorem  : 

Any  section  of  a  rigid  circular  cone  made  by  a  jjlaiie  not  pass- 
ing through  the  vertex  is  a  circle,  an  ellij^se,  an  hyperbola,  or 
a  parabola. 

Let  a  right  circular  cone  with  vertex  0  and  base  AB  be  cut 
by  a  plane,  not  passing  through  the  vertex ;  to  prove  the 
section  a  circle,  ellipse,  hyperbola,  or  parabola. 

If  the  cutting  plane  is  parallel  to  the  base,  the  section  is 
a  circle  by  Solid  Geometry. 

If  the  plane  is  not  parallel  to  the  base,  let  the  plane  OAB 
be  taken  perpendicular  both  to  the  base  of  the  cone  and  to  the 
cutting  plane.  This  plane  necessarily  contains  the  axis  of  the 
cone,  and  hence  passes  through  the  centre  of  each  circular 
section  of  the  cone.  Let  MN  be  the  intersection  of  OAB  and 
the  cutting  plane.     We  have  then  to  distinguish  three  cases. 

Case  I.      ll'/u'/i  MN  7)ieets  both  of  the  elements  OA  a?id  OB. 

Take  C,  Fig.  68,  the  middle  point  of  M  N  and  Q  any  other 
point,  and  pass  planes  parallel  to  the  base  through  C  and  Q. 


152 


PLANE   ANALYTIC   GEOMETRY. 


Fig.  68. 


The  new  sections  are  circles  of  which  the  lines  FG  and  H  K 
are  diameters.  QP,  the  intersection  of  the  planes  FG  and 
MN,  is  perpendicular  to  the  plane  OAB  and  hence  perpen- 
dicular to  the  lines  FG  and  MN  by  Solid  Geometry.  For 
similar  reasons,  CR  is  perpendicular  to   HK  and  MN. 

Now,  since  the  sections  FG  and  H  K  are  circles  of  which  the 
lines  FG  and  HK  are  diameters, 

QP'=FQ.QG, 
and  CR'^HC.CK. 

But,  from  similar  triangles, 

FQ^MQ 
HC~  MC 

QG^QN 
CK       CN" 


and 


QP- 


IQ.QN 


CR'      MC.CN 


(1) 


THE   CONIC    SECTIONS.  153 


Let  us  place 

CR  =  b, 

and 

MC  =  CN-=a. 

We  have,  then, 

QP'      MQ.QN 
b-              a- 

(2) 

Finally,  in  the 
nates,  and  M  N  as 

plane   MN,  take  C  as  an 
the  axis  of  x. 

origin  of 

coordi- 

Then,  for  the  point 

P, 

MQ: 

QH-- 

QP=y, 
CQ  =  x, 

=  MC  +  CQ  =  a  +  x, 
=  CN-CQ  =  a-x. 

Hence  equation  (2) 

becomes 

y^       a-  —  X- 
b^  ~      a'     ' 

or 

2              2 

-  +  ^   =1 

Therefore,  the  section  MN  is  an  ellipse. 

Case  II.     IV/ien  MN  meets  one  of  the  elements,  Ok,  produced. 

Let  the  construction  be  made  exactly  as  in  Case  I.     Then, 
from  the  section  FG,  Fig.  69, 

QP'=FQ.QG. 
But  from  similar  triangles, 

FQ^MQ 
KG       MC' 

and  QG_NQ. 

^"""^  HC~CN 

Hence  Q^'=  MCXN  ■^^*^- ^^^-  ^^^ 

Now  place  MC  =  CN=a, 

and  KC.HC=b2. 


164 


PLANE   ANALYTIC   GEOMETRY. 


/       n^  \ 
Ae— T 


Pig.  69. 
Equation  (1)  becomes 

QP'^%^  MQ.NQ. 

di 


(2) 


If,  finally,  in  the  plane  MN  we  take  C  as  the  origin  of  coor- 
dinates and  M  N  as  the  axis  of  x,  then 

MQ=MC  +  CQ  =  a+ X, 
NQ=  CQ-CN  =  x-a, 
and  equation  (2)  becomes 

y2  =  ^(x^'-a^, 


or  —  —  —  =  1. 

a"'       b 

Hence,  the  section  is  an  hyiierhola. 


THE   CONIC   SECTIONS. 


155 


Case  III.      When  M  N  is  parallel  to  one  of  the  elements  OA. 

0 


Fig.  70. 

Take  Q,  Fig.  70,  any  point  on  the  line  MN,  and  N,  the  ex- 
tremity of  M  N,  and  pass  planes  parallel  to  the  base  of  the 
cone,  as  before. 

Then  QP'=FQ.QG. 

But  FQ=HN, 

hn~oh' 


and 


Q^'-^-NQ. 


(1) 


Now  take  N  as  the  origin  of  coordinates,  and  MN  as  the 
axis  of  X. 

Then  QP  =  y, 

__  NQ=x, 

HN' 
and  TTTT)  since  it  is  constant,  may  be  represented  by  4p. 
OH 


*156  PLANE   ANALYTIC    GEOMETRY. 

Equation  (1)  is  then 

and  the  section  is  a  parahola. 

Limiting  Cases.  If  we  consider  now  the  case  in  which 
the  cutting  plane  passes  through  the  vertex,  we  have  from 
Solid  Geometry,  the  theorem  : 

Any  section  of  a  right  circular  cone,  made  hy  a  plane  j^assinff 
through  the  vertex,  is  a  point,  two  intersecting  straight  lines, 
or  two  coincident  straight  lines. 

In  particular,  if  the  plane  which  cuts  out  a  circle  or  an 
ellipse  be  made  to  approach  the  vertex,  the  circle  or  ellipse 
grows  constantly  smaller,  until  finally  it  becomes  a  point. 

Hence,  the  point  is  a  limiting  case  of  the  ellip)se  or  circle. 

If  the  plane  which  cuts  out  an  hyperbola  approaches  the 
vertex,  the  hyperbola  approaches  two  straight  lines. 

Hence,  two  intersecting  straight  lines  form  a  limiting  case  of 
the  hyperbola. 

If  the  plane  which  cuts  out  a  parabola  is  made  to  pass 
through  the  vertex,  the  plane  becomes  a  tangent  plane,  inter- 
secting the  cone  in  two  coincident  straight  lines. 

Hence,  two  coinciderit  straight  lines  form  a  limiting  case  of 
the  parabola. 

Finally,  if  the  vertex  of  the  cone  is  made  to  recede  indefi- 
nitely from  the  base,  the  cone  approaches  a  cylinder  as  a 
limit,  and  the  parabolic  section  approaches  two  parallel 
straight  lines. 

Hence,  tioo  parallel  straight  lines  form  a  limiting  case  of  the 
parabola. 

We  have  obtained  thus  all  the  loci  discussed  in  the  previ- 
ous articles,  except  the  "  imaginary  ellipse  "  and  "  imaginary 
straight  lines."  Of  course,  these  loci  cannot  be  obtained 
from  a  real  cone,  since  they  have  no  real  existence.  We 
agree  to  class  them  with  the  conic  sections,  because  of  the 
similarity  of  their  equations. 


THE    CONIC    SECTIONS. 


167 


74.     Polar  Equation  of  the  Conic  Section,  the  Focus  being 
the  Pole. 

The  equation  of  the  conic  section, 

y^+(x  — 2p)^  =  eV, 
may  be  transformed  to  the  focus  as  origin,  by  placing 

y  =  y^ 

x  =  x'  +  2p. 
It  then  becomes 

y'^'  +  x'^^  =  e^'(x'  +  2p)l 

By  [7],  §  10,  to  transform  to  polar  coordinates,  we  have  to 

place 

x'  =  r  cos  6, 
y'  =:  r  sin  6. 
There  results 

r^  =  e-  (r  cos  6 -\-  2py, 
or  r  =  ±  e  (r  cos  6  +  2p). 

Whence,  by  separating  the  two  values  of  r, 
2p  ^_  2p 


1  —  6  cos 


1  +  e  cos 


EXAMPLES. 

Determine   the    nature   and    equations    of    the   following 
conies  : 


1.  Focus,       (4,  0) 

2.  -       (-5,0) 

3.  "  (0,  2) 

4.  -  (1,  2) 

5.  -        (3,-4) 

6.  -       (-2,3) 


directrix,  x  =  0 

''  x  =  0 

''  y=0 

'^  X  =  0 

u  y  =  1 


e  —  J. 
e  =  2. 

6  =  1. 

P  =  1 
e  —  3. 

e  =  «. 


X  =  —  4;     6  =  1. 


Find  the  equations  of  the  following  parabolas   from  the 
given  data  : 

7.  Focus,       (0,  1) ;    vertex  (0,  0). 

8.  -       (0,-1);        -       (0,0). 


158 


PLANE   ANALYTIC    GEOMETRY. 


9. 

Focus, 

(-3,0) 

vertex, 

(0,  0). 

10. 

a 

(4,0) 

a 

(0,  0). 

11. 

ii 

(0,0) 

li 

(0,  2). 

12. 

u 

(-2,3) 

ii 

(3,  3). 

13. 

a 

(1,2) 

(I 

(1,  1)- 

Find  the  equations  of  the  following  ellipses,  the  axes  of 
coordinates  being  parallel  to  the  axes  of  the  curves  : 

14.  a  =  3  ;  b  =  1 ;  centre,       (0,  0). 

(0,  0). 

(1,  2). 


15.  a  —  2  j  b  —  ^  ; 

16.  a  =  l;  b  =  i; 


17.  a  =2;  b  =  4;        "      (-3,  0). 
-18.  a-f;  b  =  i;        "(-2,-1). 

19.  Vertices,  (±  5,  0)  ;  one  focus,  (3,  0). 

20.  Foci,  (±  3,  0)  ;  directrices,  x  =  ±  4. 

21.  Vertices,  (—  1,  0),  (5,  0)  ;  one  focus,  (0,  0). 

22.  Minor  axis,  6  ;  one  focus,  (—  4,  0)  ;  centre,  (0,  0). 

23.  Minor  axis,  12  ;    e  =  y%  ;  centre,  (0,  0)  ;    minor  axis 
along  OX. 

24.  Major   axis,  6  ;    e  =  ^  ;    centre,    (0,   0)  ;    major   axis 
along  OX. 

25.  Centre,  (0,  0) ;  passing  through  (3,  4)  and  (1,  5)  ;  major 
axis  along  OX. 

Find  the  equations  of  the  following  hyperbolas,  the  axes 
of  coordinates  being  parallel  to  the  axes  of  the  curves  : 

26.  a  =  1  ;  b  =  6 ;  centre,  (0,  0);  transverse  axis  along  OX. 

27.  a  =  3  ;  b  =  4-;  centre,  (0,  0) ;  transverse  axis  along  OY. 

28.  a  =  §  ;  b  =  i  ;  centre,  (0,  0);  transverse  axis  along  OX. 

29.  a  =  1 ;  b  =  2 ;  centre,  (1,  —  2) ;  transverse  axis  parallel 
to  OX. 

30.  a  =  Vj  ;  b  =  V|^ ;  centre,  (2,  3)  ;  transverse  axis  par- 
allel to  OY. 

31.  Vertices,  (±  4,  0)  ;  e  =  f . 

32.  Foci,  (±  7,  0)  ;  directrices,  x  =  ±  5. 


THE    CONIC    SECTIONS.  159 

33.  a  =  5;  centre,  (0,0);  e  =  1^  ;  transverse  axis  along 
OX. 

34.  Centre,  (0,  0)  ;  passing  through  (1,  3)  and  (—  2,  —  4)  ; 
conjugate  axis  along  OX. 

35.  Centre,  (0,  0) ;  passing  through  (1,  1)  ;  transverse  axis 
along  OX  ;  asymptotes  having  slopes  ±  |. 

36.  Passing  through  (1,  4)  ;  having  same  asymptotes  as 
9y2  ^  x^  ==  8. 

Determine  the  nature  of  the  following  conies.  If  the  conic 
is  a  parabola,  find  (1)  vertex,  (2)  focus,  (3)  axis,  (4)  directrix. 
If  the  conic  is  an  ellipse  or  an  hyperbola,  find  (1)  centre,  (2) 
equation  of  major  axis,  (3)  lengths  of  semi-axes,  (4)  ver- 
tices, (5)  eccentricity,  (6)  foci,  (7)  directrices.  If  the  conic 
is  an  hyperbola,  find  also  (8)  the  asymptotes. 

37.  4x-  +  49y2  =  196. 

38.  3x-  +  2y-^  =  l. 

39.  x2==-7y. 

40.  ^-t=:-l. 

16       9 

41.  y'^  — 16x  =  0. 

42.  2x2-3y^==6. 

43.  x^^lSy. 

44.  36x2  _  isy2  _|_  36x  +  24y  -  35  =  0. 

45.  x2  — 4x  — lly  — 7==0. 

46.  3x2  +  y2  — 6x-6y  +  9  =  0. 

47.  4y-2-4y  +  6x  +  3  =  0. 

48.  x2  +  2y2  +  6x  +  8y  +  4  =  0. 

49.  12x2  _  igy2  _  3gx  _  i2y  -f  31  =  0. 

50.  3x2  _  7y2  _  6x  —  28y  —  46  =  0. 

51.  9x-  +  25y2  —  36x  +  50y  —  164  =  0. 

52.  y  =  (x  -  2)  (x  -f  3). 

53.  y2  =  (x  +  3)  (4  -  x). 

54.  y2  =  (x  +  3)  (x  -  4). 


160  TLANE   ANALYTIC    GEOMETRY. 

55.  Find  the  equation  of  the  ellipse  in  terms  of  a  and  b, 
when  the  origin  is  at  the  vertex,  the  major  axis  lying  along 
OX. 

56.  Find  the  equation  of  the  hyperbola  in  terms  of  a  and 
b,  when  the  origin  is  at  the  vertex,  the  major  axis  lying 
along  OX. 

57.  Prove  that  in  tlie  ellipse  or  hyperbola  the  squares  of 
the  ordinates  of  any  two  points  are  to  each  other  as  the 
products  of  the  segments  of  the  transverse  axis  made  by  the 
feet  of  these  ordinates. 

58.  Prove  that  in  the  ellipse  or  hyperbola,  the  semi-conju- 
gate axis  is  a  mean  proportional  between  AF  and  FA'. ' 

59.  Prove  that  in  the  ellipse  or  hyperbola,  the  semi-latus 
rectum  is  an  harmonic  mean  between  AF  and  FA'. 

60.  A  point  moves  so  that  the  difference  of  its  distances 
from  (a,  b)  and  (a,  —  b)  is  always  equal  to  2c.  Find  its 
locus. 

61.  A  point  moves  so  that  the  sum  of  its  distances  from 
(a,  b)  and  (a,  —  b)  is  always  equal  to  2c.     Find  its  locus. 

62.  A  line  of  constant  length  k  moves  with  its  extremities 
on  the  two  axes  of  coordinates.  Find  the  locus  described  by 
any  point  of  the  line. 

63.  Show  that  -^ Hj  +  r^ — r^  =  1,  where  k  is  an  arbitrary 

X        y 
parameter,  represents  ellipses  confocal  to  -^  +  fi  ^  1  when 

k'^  <  b^   or   k-  >  a^,  and   represents    hyperbolas   confocal   to 

X  V 

—  +  f-„  ^  1  when  k"  >  b"  but  <  a^,  a^  being  considered  greater 
a'      b" 

than  bl 

64.  Show  that  if  the  focus  is  taken  upon  the  directrix,  the 
conic  section  becomes  one  of  the  limiting  cases. 


CHAPTER    VII. 

TANGENT,  NORMAL,  AND    POLAR. 

75.  We  propose,  in  this  chapter,  to  discuss  three  straight 
lines,  which  we  have  found  to  be  of  importance  in  the  case  of 
the  circle  :  namely,  the  tangent,  the  normal,  and  the  polar. 
The  discussion  will  be  almost  exactly  like  that  of  the  chapter 
on  the  circle,  but  we  will  proceed  without  direct  reference  to 
that  chapter,  both  for  the  sake  of  the  review  of  important 
properties,  and  also  because  of  the  greater  generality  in  our 
treatment, 

76.  Tangent. 

Beginning  with  the  tangent,  we  have  the  definition  : 

A  tangent  at  any  point  of  a  curve  is  the  straight  line  ivhich 
the  secant  through  that  point  and  any  second  p)oint  of  the  curve 
approaches  as  a  limit,  as  the  second  point  approaches  the  first 
along  the  curve. 

In  accordance  with  this  definition,  the  slope  of  a  tangent 
may  be  found  when  the  point  of  contact  is  known,  by  first 
finding  the  slope  of  the  secant,  and  then  passing  to  the  limit. 

Let  Pi(xi,  yi)  and  P2(x2,  ya),  Fig.  71,  be  two  neighboring 
points  upon  a  conic,  the  equation  of  which  is 

Ax'^  +  By-  +  2  Gx  +  2  Fy  +  C  =  0. 

Draw  the  secant  PiPo  and  the  tangent  PjT,  and  let  the  angle 
RP,P2  =  ^,  and  RPiT  =  <^. 
Then 

tan^^^y^:^- 

Xo  —  X, 


162 


PLANE    ANALYTIC    GEOMETRY. 
Y 


Fig.  71. 

As  P2  approaches  P^  y.2  approaches  yi,  Xg  approaches  Xi,  and 

Si  —  Vi  0 

the  fraction —  approaches  the  indeterminate  form  -. 

X2  — xi     ^^  0 

Hence  it  is  impossible  to  find  the  limiting  value  of  tan  0 
from  this  fraction  directly.  But  we  can  find  another  expres- 
sion for  tan  6,  by  considering  the  fact  that  both  Pj  and  Pg  are 
on  the  conic. 

We  have,  namely, 

Axi^  +  Byi^  +  2  Gxi  +  2  Fyi  +  C  =  0,  (1) 

and  Ax2-+ By2-  +  2Gx2  +  2Fy2  +  C==0.  (2) 

By  subtracting  (1)  from  (2),  we  obtain 
A(x2^-Xi^)+B(y2^-yi^)  +  2G(x2-x0  +  2F(y2-yi)=0, 
and,  by  factoring, 
(x2-xO[A(x2+xO  +  2G]  +  (y2-yO[B(y2  +  yi)  +  2F]  =  0. 

Whence 

y2-yi^     A(x2  +  xO  +  2G 

xo-xi  B(y2  +  yO+2F 


TANGENT,   NORMAL,    AND   POLAR.  163 

Hence,  we  have 

t„„fl-      A(x,  +  x,)+2G 
'''"^—  B(y.  +  y,)+2F 

If  now  P2  approaches  Pi,  tan  6  approaches  tan  <^,  and 

B(y.  +  yi)+2F''PP'°^^^^'       By^  +  F 
Hence,  by  the  theory  of  limits, 

.      ^  Axi  +  G 

'"^^  =  -By7+7- 

The  equation  of  the  tangent  is,  therefore,  by  [13],  §  33, 

But  this  equation  may  be  placed  in  a  much  more  conven- 
ient form.  Let  us  first  clear  of  fractions  and  transpose. 
There  results 

Axix  —  Axi^  +  By^y  -  By^^  +  Gx  —  Gxi  -f-  Fy  —  Fy^  =  0. 
Whence,  by  adding  (1), 

AxiX  +  ByiY  +  G  (X  +  Xi)  +  F  (y  +  yO  +  C  =  0.     [34] 

This  equation  of  the  tangent  may  be  readily  remembered 
by  comparing  it  with  the  equation  of  the  conic.  It  will  be 
seen  that,  as  in  the  case  of  the  circle,  ive  have  only  to  replace 
x^  arid  y^  in  the  equation  of  the  conic  by  XjX  and  \ji\j,  respectively, 
and  to  replace  2x  and  2y  6y  x  -(-  Xj  and  y  +  yi,  respectively,  in 
order  to  obtain  the  equation  of  the  tangent. 

The  following  are  special  cases  : 

(1)  Parabola,  y-  =  4px  ;  tangent,  yiy  =  2p  (x  +  x,)- 

(2)  Ellipse,  ^.  +  ^2=1;  tangent,  ^  +  M  _  i. 

v\/  Yy\/\/ 

(3)  Hyperbola,  -  —  ^-  =  1 ;  tangent,  Tir  —  xf  =  1- 


a' 


b^          '         «       '   a-         b^ 


Ex.     Find  the  equation  of  the  tangent  to  the  hyperbola 
2x2  -  3y2  +  4x  +  G  =  0, 


164  PLANE    ANALYTIC    GEOMETRY. 

passing  through  the  point  (1,  —  2).  Since  (1,  —  2)  satisfies  the  equation 
of  the  hyperbola,  it  is  the  point  of  contact.     Hence  the  tangent  is 

2x  (1)  -  3y  (-  2)  +  2  (X  +  1)  +  6  =  0, 

or  2x  +  3y  +  4  =  0. 

For  some  purposes,  it  is  convenient  to  be  able  to  find  a 
tangent  with  a  given  slope,  the  point  of  contact  being  un- 
known. In  such  cases  the  method  used  in  §  18  and  §  53  is 
sufficient  to  solve  the  problem.  We  call  m,  namely,  the 
given  slope.    Then  the  equation  of  the  tangent  is  of  the  form 

y  =  mx  +  c, 

where  c  is  to  be  so  determined  that  the  points  of  intersection 
of  the  curve  and  the  tangent  shall  be  coincident.  The 
student  may  verify  the  following  results  for  special  cases  : 

(4)  For  the  parabola  y^  :=  4px,  there  is  one  tangent  with 

slope  m  ;  namely,  y  =  mx  +  ^  • 

x         V 

(5)  For   the   ellipse  "i  +  f^  =  Ij  there  are  two  tangents 

d.        D 


with  slope  m  ;  namely,  y  =  mx  ±  Va^m^  +  b^. 

y2  y2 

(6)    For  the  hyperbola  —  —  ui  ^=  !>  there  are  two  tangents 
a       o 

with  slope  m  ;  namely,  y  =  mx  ±  Va^m^  —  b^ 

Ex.     Find  the  tangents  to  the  conic  2x2  +  3y2  —  4x-|-6y-|-4=0, 
which  are  parallel  to  the  line  4x  —  2y  +  3  ==  0. 

Here,  m  =  2.     Hence  the  tangent  is  y  =  2x  +  c. 
Substituting  in  the  equation  of  the  curve,  we  have,  for  the  abscissas  of 
the  points  of  intersection,  the  equation 

14x2  +  (12c  +  8)  X  +  3c2  +  6c  +  4  =0. 
This  equation  has  equal  roots,  if 

(12c  +  8)2  =  56  (3c2  +  6c  +  4), 
whence  c  =  -  3  ±  i  V21. 

Hence  the  tangents  are  y=:2x— 3±i  V21. 


TANGENT,   NORMAL,    AND   POLAR.  165 

77.    The  Normal. 

The  normal  to  a  curve  at  any  point  is  the  straight  line  pass- 
ing through  that  point  and  perpendicular  to  the  tangent  at  that 
point. 

To  find  the  equation  of  the  normal,  it  is  most  convenient 
first  to  find  the  equation  of  the  tangent.  Then,  if  m  is  the 
slope  of  the  tangent,  the  equation  of  the  normal  is 

y-yi  =  --(x-xi)- 

The  following  are  special  cases  : 

(1)  For  the  parabola  y-  =:  4px,  the  normal  is 

yi  /  \ 

y-yi  =  -2^(x-xO. 

9  9 

X"       y 

(2)  For  the  ellipse  ~^  +  f^  =  !>  the  normal  is 

(3)  For  the  hyperbola  -^  —  ^^  =  1,  the  normal  is 

Ex.     Find  the  equation  of  tlie  normal  to  the  parabola 
x2  +  2x  +  3y  -  9  =  0, 
at  the  point  (—  2,  3). 

The  equation  of  the  tangent  at  this  point  is 

(-  2)  X  +  (X  — 2)  +  I  (y  +  3)  -  9  =  0, 
or  2x  -  3y  +  13  =  0. 

Therefore,  the  equation  of  the  normal  is 

y-3  =  -  j!(x  +  2), 
or  3x  +  2y  =  0. 

If  the  slope  of  the  normal  is  given,  the  normal  may  be 
found  as  in  the  following  example. 

Let  it  be  required  to  find  the  normals  to  the  conic 
2x2  +  3y2  _  4x  +  6y  +  4  =  0, 
which  are  parallel  to  3x  —  2y  +  0  =  0. 


166  PLANE   ANALYTIC   GEOMETRY. 

Calling  m  the  slope  of  the  normal,  we  have,  at  once,  m  =  |.  If  we 
can  find  the  point  where  the  normal  cuts  the  curve,  we  can  solve  the 
problem.  Let  us  call  (xi,  yi)  this  point.  Then  the  equation  of  the 
tangent  is,  by  [34],  §  76, 

2xix  +  3yiy  -  2  (x  +  Xi)  +  3  (y  +  yi)  +  4  =  0, 

whence,  if  mi  is  the  slope  of  the  tangent, 

_      2x1-2      T,  ,  1 

mi-  — q.,    I   q-     But  mi  = ; 

oyi  +  3  m 

2xi  -  2  _  2 
•'■  3yi  +  3~3' 
or  xi  —  yi  =  2. 

But  (xi,  yi)  is  on  the  conic ;  hence 

2xi2  +  3yi2  -  4xi  +  6yi  +  4  =  0. 
These  two  equations  are  sufficient  to  determine  (xi,  yi).     We  find 
xi  =  l±^V5, 
yi  =  -  1  ±  ^  Vs. 
Hence  there  are  two  normals,  their  equations  being 

y  +  l:F^V5=|(x-l:F|  VB), 
or  y  =  |x-|±xVV5. 

78.  Pole  and  Polar, 

If,  through  a  fixed  iwint  P^  any  number  of  secants  he  drawn 
to  a  conic  section,  and  tangents  to  the  latter  he  drawn  at  the 
points  of  intersection  of  each  secant  with  the  co7iic,  then  each 
pair  of  tangents  will  intersect  upon  a  straight  line,  called  the 
polar  of  Pj.  Reciprocally  Pi  is  called  the  pole  of  the  sti'aight  line. 

Let  Ax2+ By2  +  2  Gx  +  2  Fy  +  C  =  0 

be  the  equation  of  the  conic  section,  and  (xi,  yi)  the  coordi- 
nates of  the  point  Pi.  Then  if  Pi  lies  on  the  conic  section,  it 
follows  from  the  definition  that  the  polar  of  Pi  is  simply  the 
tangent  at  Pi,  as  shown  in  Fig.  72. 

In  this  case,  then,  the  equation  of  the  polar  is 

Axix  +  Byiy  +  G  (x  +  Xi)  +  F(y  +  y,)  +  C  =  0. 

The  above  case  is  unimportant ;  the  polar  has  real  im- 
portance when  the  point   Pi  is  not  on  the  conic.     In  this 


TANGENT,   NORMAL,    AND   POLAR. 


167 


FlO.  72. 


case  we  may  draw  for  the  ellipse  either  one  of  Figs.  73  and  74, 
and  similar  figures  for  the  parabola  and  the  hyperbola. 


Fig.  73. 

Take  any  secant  PiN  intersecting  the  conic  at  N  and  M, 
and  let  (xa,  y^)  and  (xs,  ys)  be  the  unknown  coordinates  of  N 
and  M.  The  equations  of  the  tangents  at  N  and  M  are, 
respectively,  by  [34],  §  76, 

Ax.,x  +  By,y  +  G  (x  +  x,)  +  F  (y  +  y,)  +  C  =  0,  (1) 

Ax,x  +  By,y  +  G  (x  +  X3)  +  F  (y  +  y,)  +  C  -  0.  (2) 


168 


PLANE   ANALYTIC    GEOMETRY. 


Fig.  74. 

The  coordinates  of   R,  the  point  of  intersection  of  these 
tangents,  satisfy  both  of  these  equations  and  hence  satisfy 
their  difference 
A(x2-x3)x  +  B(y2-y3)y  +  G(x2-X3)  +  F(y2-y3)  =  0.  (3) 

This  equation  contains  the  unknown  coordinates  (xg,  y^) 
and  (xg,  y3),but  does  not  contain  the  given  coordinates  (xi,  yi), 
a  defect  which  may  be  remedied  by  remembering  that  Pj,  M, 
and  N  lie  upon  a  straight  line,  and  that  therefore 

X  2  '^~  X  3  X  J  ~~~*  X  2 

y2  — ys     yi  — 72' 

Substituting  in  (3),  we  have 

A  (xi  -  X2)  x  +  B  (yi  -  y^)  y  +  G  (x^  -  x^)  +  F  (y^  -  y^)  =  0, 
and  adding  (1),  we  have,  finally, 

AxiX  +  Byiy  +  G  (x  +  Xi)  +  F  (y  +  yi)  +  C  =  0.     [35] 

This  is  an  equation  satisfied  by  the  coordinates  of  R.  It 
contains  the  coordinates  of  the  given  point  Pi,  and  not  the 
coordinates  of  M  and  N.  Hence  the  result  would  be  un- 
changed if  we  had  used  any  other  secant  passing  through  Pj. 
Moreover,  equation  [35]  is  an  equation  of  the  first  degree, 


TANGENT,    NORMAL,    AND    POLAR.  169 

and  therefore  represents  a  straight  line.  Hence  the  theorem 
at  the  beginning  of  the  article  is  true. 

It  is  to  be  noticed  that  the  form  of  the  result  is  the  same 
as  that  obtained  when  the  point  Pi  lay  on  the  conic.  In 
other  words,  the  equation  of  the  'polar  is  identical  in  form  xvith 
the  equation  of  the  tangent. 

Hence,  if  the  point  (xj,  yi)  is  given  and  we  form  equa- 
tion [35],  it  will  not  be  known  whether  this  is  the  tangent  or 
the  polar  until  we  examine  the  position  of  the  point  (xj,  yj). 
If  (xi,  yi)  lies  on  the  conic  we  have  the  tangent ;  otherwise, 
we  have  the  polar. 


Ex.     Given  the  ellipse 


9^4         ' 


and  the  point  (4,  —  5).     The  coordinates  (4,  —  5)  do  not  satisfy  the 
equation 


,,+i  =  '- 


Therefore,  the  line 

is  the  polar  of  the  point  (4,  —  5) 


9        4 
4x      5y 


9        4=1 


79.     To  find  the  Tangents  from  a  Point  not  on  the  Conic 
Section. 

Let  Pi(xi,  yi)  be  a  point  not  on  the  conic  section. 


FlO.  75. 


170  PLANE    ANALYTIC    GEOMETRY. 

Assuming  that  it  is  possible  to  draw  tangents  from  Pi  to  the 
conic,  let  us  call  (xj,  y^  the  coordinates  of  the  point  of  tan- 
gency,  Q,  of  any  tangent.  If  it  is  not  possible  to  draw  the 
tangents,  that  will  be  shown  in  the  result  by  the  fact  that 
X2  and  y2  have  imaginary  values. 

The  equation  of  the  tangent  QPi  is,  by  [34],  §  76, 

Ax.,x+  By,y  +  G  (x  +  x,)  +  F  (y  +  y^)  +  C  =  0.  (1) 

Since  this  tangent  passes  through  the  point  P^,  the  coordi- 
nates (xi,  yi)  satisfy  the  equation  (1),  and  we  have 

Ax^xi  +  By^yi  +  G  (x^  +  x,)  +  F  (y^  +  y^)  +  C  =  0.       (2) 

Now  consider  the  polar  of  Pj.     Its  equation  is,  by  [35], 

Axix  +  Byiy  +  G  (x  +  xO+  F  (y  +  yO  +  C  =  0,  (3) 

and  equation  (2)  shows  that  the  coordinates  X2,  yo  satisfy  (3), 
that  is,  that  the  point  Q  lies  upon  the  polar  of  Pj.  We  see, 
therefore,  that  tangents  from  a  point  Pi  to  a  conic  touch  the 
conic  in  the  points  where  the  latter  is  cut  by  the  polar  of  Pi- 
Hence,  to  find  the  tangents  from  a  2>oint  Pi  to  a  conic,  proceed 
as  folio ivs  : 

First,  find  the  polar  of  Pi.  llien,  find  the  2Joi?its  of  inter- 
section of  the  polar  and  the  conic.  Fhially,  find  the  tangents  at 
these  points. 

Ex.     Find  the  equations  of  the  tangents  to  the  ellipse 
4x2  +  y2  +  24x  -  2y  +  17  =  0, 
from  the  point  (—  |,  ^-f).     The  polar  of  the  point  is 
2x  +  y  —  1  =  0. 
Solving  the  two  equations,  we  find  that  the  polar  intersects  the  curve 
in  the  points  (—  1,  3)  and  (—  2,  5).     The  tangents  at  these  points  are 

4x  +  y  +  1  =0, 
and  X  +  y  —  3  =  0. 

Since  the  polar  is  a  straight  line,  it  may  have  any  one  of 
three  relations  to  the  curve. 


TANGENT,    NORMAL,    AND    POLAR.  171 

(1)  It  may  be  tangent  to  it. 

(2)  It  may  cut  it  in  two  real  points. 

(3)  It  may  not  intersect  it  at  all. 

The  first  case  is  possible  only  when  the  pole  is  itself  on  the 
conic  ;  the  last  two  occur  when  the  pole  is  not  on  the  conic. 
In  case  (2),  by  applying  the  rule,  we  find  two  tangents  from 
the  pole.  The  polar  is  then  the  chord  connecting  the  points 
of  contact  of  the  tangents  from  the  pole  or  tlie  so-called 
chord  of  contact.  In  case  (3),  by  the  application  of  the  rule, 
we  see  that  no  real  tangents  can  be  drawn. 

From  this  follows  the  theorem  : 

From  a  2Joint  not  on  a  conic,  either  two  or  no  tangents  can  he 
drawn  to  the  conic. 

We  may  accordingly  separate  the  points  of  a  plane  into 
two  classes,  those  outside  and  those  inside  the  conic  ;  the 
latter  being  those  from  which  no  tangent  can  be  drawn. 

And  now,  summing  up,  we  may  say  : 

According  as  a  point  Pi  lies  (1)  without,  (2)  on,  (3)  within  a 
conic,  the  polar  of  P^  is 

(1)  the  chord  of  contact, 

(2)  the  tangent, 

(3)  a  straight  line  not  intersecting  the  conic. 

If  it  is  questioned  whether  a  point  lies  within  or  without  a 
conic,  the  question  may  be  answered  by  determining  the  posi- 
tion of  the  polar;  unless,  as  is  often  the  case,  the  question 
may  be  more  quickly  solved  by  inspection. 


80.     Properties  of  Poles  and  Polars. 

Two  of  the  most  important  properties  of  polars  are  the  fol- 
lowing : 

1.  If  Pa  lies  upon  the  polar  of  Pi,  then  reciprocall//  Pi  lies 
tipon  the  polar  of  P^- 


17,2  PLANE   ANALYTIC    GEOMETRY. 

Let  the  coordinates  of  Pi  be  (xi,  yj),  those  of  Pa  be  (xa,  yo)- 
The  polar  of  Pi  is 

Axix  +  Byiy  +  G  (x  +  x^)  +  F  (y  +  y,)  +  C  =  0,       (1) 

and  that  of  Pj  is 

Ax,x+By,y  +  G  (x  +  x,)  +  F  (y  +  y,)  +  C  =  0.        (2) 

The  condition  that  Pj  shoukl  lie  on  (1)  is 

Axix,+  Byiy,+  G  (x^  +  x^)  +  F  (y,  +  yO  +  C  =  0  ; 

but  this  is  exactly  the  condition  that  Pi  should  lie  on  (2). 
This  establishes  the  theorem.  This  theorem  may  be  other- 
wise stated  as  follows : 

The  polars  of  all  i^oints  lying  in  the  same  straight  line  pass 
through  the  same  jjoint,  namely  the  pole  of  that  straight  line. 

With  the  aid  of  this  theorem,  theorems  similar  to  those  in 
§  60  may  be  readily  established  and  many  others,  the  discus- 
sion of  which  lies  outside  the  range  of  this  book. 

2.  Any  secant  passing  through  Pi  is  divided  harmonically  by 
the  conic  and  the  j^olar  of  Pi. 


Fig.  76. 


Let  PiN  be  any  secant  through  Pi,  and  let  M  and  N  be 
the  points  where  this  line  is  cut  by  the  conic,  and  Q  the  point 
where  it  is  cut  by  the  polar  of  Pi.  We  are  to  prove  that  the 
line  MN  is  divided  harmonically;  i.e.,  that  it  is  divided 
externally  and  internally  in  the  same  ratio.     We  must  have 


TANGENT,    NORMAL,    AND    POLAR.  173 

then  1^=^. 

PiN       QN' 

1  f  11  t3^      2.P1N.P1M 

whence  follows  P,Q  =  ^^ttt  ,    ^  .  -  • 

Let  us  take  the  point  Pi  as  the  origin  of  coordinates  and 
consider  the  axes  of  coordinates  drawn  parallel  to  the  axes  of 
the  conic.     Then  the  equation  of  the  conic  is 

Ax2+By2  +  2Gx  +  2  Fy  +  C  =  0,  (1) 

and  the  equation  of  the  polar  of  Pi,  found  by  placing  Xi  =  0 
and  yi  =  0  in  [35],  is 

Gx+Fy  +  C=:0.  (2) 

Let  us  now  introduce  polar  coordinates,  using  [7],  §  10. 
Equations  (1)  and  (2)  become  resj)ectively 

r=^  (A  cos^  ^  +  B  sin^  ^)  +  r  (2  G  cos  ^  +  2  F  sin  ^)  +  C  =  0,  (3) 

and  r  (G  cos  e  +  F  sin  ^)  =  —  C.  (4) 

The*  roots  of  (3)  are  the  distances  PiM  and  PiN.  Hence, 
by  the  theory  of  quadratic  equations, 

C 


P,N.  PiM  = 
and  P^N  +  PiM=- 


A  cos='^+  B  sin^^' 
2  G  cos  e  +  2  F  sin  6 


A  cos^  ^  +  B  sin^  6 
The  value  of  r  in  (4)  is  the  distance  PiQ  ;  hence 

p  0=r  ~  ^ 

'^      G  cos  ^  +  F  sin  e 
These  results  give,  at  once, 

pp,^2P.N.P.M 
'^      PiN  +  PiM' 
as  was  to  be  proved. 

This  property  is  sometimes  taken  as  the  definition  of  the 
polar. 


174  PLANE   ANALYTIC   GEOMETBY. 


EXAMPLES. 

Find  the  tangents  and  the  normals  to  each  of  the  following 
curves  at  the  given  points  : 

X      1.  9x^  +  5y2  +  36x  +  20y  +  ll=0;  (-2,1). 

2.  9x2  +  4y-  +  36x  +  24y  =  0;  (0,0). 

3.  X-  -  3y2  +  6x  +  12y  -  9  =  0  ;  (0,  3). 

4.  y2  -  6y  -  8x  -  31  =  0  ;  (-  3,  -  1). 

5.  y^  =  9x;  (1,3). 

6.  x2  +  4y2  =  8;  (2,-1). 

7.  2x2-y='  =  14;  (3,-2). 
\^^            8.  Find  the  tangent  to  the  parabola  y^  =  4x,  which  makes 

an  angle  of  45°  with  the  axis  of  x. 
9.  Find  the  tangents  to  the  conic 

2x2  +  3y^  +  4x-6y-l=0, 
which  make  an  angle  tan~^  \  with  the  axis  of  x. 

10.  Find  the  tangents  to  the  conic  4x^  +  y-  ==  4,  which  are 
parallel  to  the  straight  line  2x  —  4y  4-5=0. 

11.  Find  the  tangents  to  the  conic  x"  —  3y^  —  4x  +  3  =:  0, 
which  are  perpendicular  to  3x  +  2y  —  1  =  0. 

12.  Find  the  tangents  to  the  conic  7x^+8y^  =  56,  which 
make  an  angle  tan~^  3  with  the  line  x  +  y  -f- 1  ^  0. 

13.  Find  a  normal  to  the  conic  x^  +  2y2  +  5y  +  2  =  0,  mak- 
ing an  angle  tan~^  \  with  the  axis  of  x. 

14.  Find  a  normal  to  the  conic  3x-  —  x  +  y  =  0,  parallel  to 
the  line  3x  +  2y  +  1  =  0. 

15.  Find  a  normal  to  the  conic  4y-  —  x-  -f  2x  —  6y  -f  20  =  0, 
perpendicular  to  the  line  3y  +  3x  +  2  =  0. 

Find  the  polars  of  each  of  the  following  points  with  respect 
to  the  given  conic,  and  also  the  points  in  which  the  polar 
intersects  the  conic  : 


16.  (-3,  4);  4y2  +  16x-4y-3=0. 

17.  (0,  0)  ;  3x2  _  2y2  _  6x  -  8y  -  6  =  0. 


TANGENT,    NORMAL,    AND    POLAR.  175 

18.  (-|,i);|  +  f  =  l. 

19.  (3,-1);  2x-  +  y--8x  +  Gy +  11=0. 

Find  the  tangents  drawn  through  the  given  points  to  the 
following  conies  : 

20.  7x2  +  3y2  +  14x  -  12y  -  36  =  0  ;  ^4^  _y_^_ 

21.  y^=|x;  (-1,-1). 

22.  2x-  -  6f  =  21 ;    (5  +  V3,  2  +  V3.) 

23.  y2  +  5x  — lOy  — 35  =  0;  (3,1). 

24.  What  is  the  slope  of  the  tangent  to  the  conic 

3x2 -18x  +  7y  + 13  =  0 

through  the  point  (1,  4)  ? 

25.  Find  the  angle  between  the  ellipse  4x2-|-9y2  =  13  and 
the  circle  x^  +  y^  ^  2  at  any  one  of  their  points  of  intersec- 
tion. 

26.  At  what  angle  does  the  straight  line  2x  —  y  +  3  =  0 
intersect  the  conic  3x-  -(-  y^  -|-  6x  —  4y  —  14  =  0  ? 

x"       v"  X       v" 

27.  Show  that  the  curves  — -.  +  ;^  =  1  and  -; =^=1  cut 

lb       i  4        5 

each  other  at  right  angles  and  are  confocal. 

28.  At  what  angles  do  the  conies 

x^  —  3y2  —  4x  —  6y  —  5  =  0  and  x^  —  4x  —  9y  —  5  =  0 
intersect  ? 

29.  Using  the  method  of  §  62,  find  a  conic  through  the 

points  of  intersection  of  ^^7^  +  ^  =  1  and  - —  j  =  1,  and  also 

through  the  point  (0,  6).     What  is  the  conic  ? 

30.  Use  the  method  of  §  62  to  find  the  equation  of  a  para- 
bola which  passes  through  the  intersections  of  the  hyperbola 
x^  —  y2  =:  1  and  the  circle  x-'  +  y^  —  4x  —  1  ==  0. 

31.  By  the  method  of  §  76,  show  that  the  tangent  to  the 
curve  y  =  x^  at  the  point  (xj,  yj  is  y  -|-  2yi  =  3xi-x. 


176  PLANE    ANALYTIC    GEOMETKY. 

32.  By  the  method  of  §  76,  show  that  the  tangent  to  y^  =  x'' 
at  the  point  (xi,  yi)  is  2yiy  +  yj^  =  3xf  x. 

33.  By  the  method  of  §  76,  show  that  the  tangent  to  xy  =  4 
at  the  point  (xi,  yi)  is  Xiy  +  yix  =  8. 

34.  Solve  Ex.  33  by  transforming  the  given  equation  to 
new  axes  bisecting  the  angles  between  the  old  axes,  finding 
the  tangent,  and  transforming  back. 

35.  Show  that  for  any  conic  section  the  polar  of  the  focus 
is  the  directrix. 

36.  Where  is  the  polar  of  the  centre  of  an  ellipse  or  hyper- 
bola with  respect  to  that  curve  ? 

37.  If  mi  is  the  slope  of  the  polar  of  a  point  P^  with 

x^      y^ 
respect  to  an  ellipse  —^-\-^^  =  l  and  mj  is  the  slope  of  the  line 

joining  Pj  to  the  centre,  show  that  mim2  =^ \-     Find  the 

SI 

similar  relation  for  the  hyperbola. 


CHAPTER    VIII. 

THE    PARABOLA  :   y2  =  4px. 

81.  We  have  seen  in  §  67  that,  if  we  take  the  axis  of  the 
parabola  as  the  axis  of  x  and  the  line  through  the  vertex  per- 
pendicular to  the  axis  of  the  parabola  as  the  axis  of  y,  the 
equation  of  the  parabola  assumes  its  simplest  form,  y-  =  4px. 
This  is  the  form  of  the  equation  we  shall  use  in  studying  the 
parabola  in  this  chapter. 

When  the  equation  is  in  this  form,  the  focus  is  tlie  point 
(p,  0),  and  the  equations  of  the  axis  and  the  directrix  are 
respectively  y  =  0  and  x  =  —  p.  Also,  according  to  §  78,  the 
polar  for  the  point  (xj,  yi)  is  yiy  =  2p  (x  +  Xi)/and  the  tan- 
gent at  the  point  (xj,  yi),  if  (xj,  yi)  is  on  the  curve,  is 
yiy  =  2p  (x  +  Xi).     If,  however,  the  tangent  is  determined  by 

its  slope  m,  its  equation,  by  (4),  §  76,  is  y  ^  mx  + 

82.  Constructions. 

Since  the  eccentricity  of  the  parabola  is  always  unity,  it 
follows  that  every  point  of  the  parabola  is  equally  distant 
from  the  focus  and  tlie  directrix.  From  this  fact  we  derive 
the  two  following  methods  of  construction  : 

(1)   By  compasses. 

Let  F  (Fig.  77)  be  the  focus  and  DD'  and  FX  be  the  directrix 
and  the  axis,  respectively,  of  the  parabola  to  be  constructed. 

The  vertex,  V,  will  be  on  the  line  FX  halfway  from  F  to 
DD'.  To  determine  other  points  of  the  parabola,  draw  a  line, 
as  AB,  parallel  to   DD'  and  to  the  right  of  V.     With  F  as  a 


178 


PLANE    ANALYTIC    GEOMETRY. 


Fig.  77. 


centre  and  a  radius  equal  to  the  distance  from  DD'  to  AB 
describe  an  arc  of  a  circle.  This  arc  will  intersect  AB  at  two 
points  P'  and  P",  which  will  be  points  of  the  parabola. 

Proceeding  in  this  way  we  may  determine  as  many  "joints 
of  the  curve  as  we  wish. 

(2)  By  a  mechanical  device. 

As  before,  F  (Fig.  78)  shall  be  the  focus  and  DD'the  direc- 
trix. ACB  is  a  triangle,  right  angled  at  C,  which  moves  with 
CB  always  coinciding  with  DD'.  A  thread  of  length  equal 
to  AC  is  fastened  to  the  triangle  at  A  and  to  the  plane  at  F. 
Then,  as  the  triangle  slides  along  DD',  the  point  of  a  pencil 
which  always  touches  AC  and  keeps  the  thread  taut  will 
describe  the  parabola. 


THE    I'AKABOLA. 


179 


Fig.  78. 


83.     Angular  Properties  of  Tangent  and  Normal. 

1.  The  tancjent  viakes  equal  angles  with  the  focal  radius 
drawn  to  the  iwint  of  contact  and  the  axis  of  tlie  -parabola. 

In  Fig.  79,  let  PjT  be  tangent  at  the  point  Pi  (xi,  yi)  to  the 
parabola  y^  ^=-  4px,  of  which  the  focus  is  F.  We  have  then  to 
prove  that  Z  FPiT  =  Z  FTPj. 

Now      FPi=  ^'^^^E^A^^iE^ 

=  Vxi^— 2pxi  +  p-  +  yi^ 

=  Vxi^  —  2pxi  +  p2  +  4pxi,       (for  yi"  =  4pxi) 

=  Vxi^  +  2pxi  -I-  p2  =  Xi  +  p. 


180 


PLANE    ANALYTIC    GEOMETRY. 


Fig.  79. 


And  TF  =  OF  -  OT  =  p  —  (—  xi)  =  p  +  xi,  for  OT  is  the 
intercept  of  the  tangent  on  OX  and  is  found  to  be  —  Xi. 

Therefore,  since  FPi  =  TF,  the  triangle  TFPi  is  isosceles 
and  the  angles  FPiT  and  FTPi,  opposite  the  equal  sides,  are 
equal. 

2.  The  normal  makes  equal  angles  with  the  focal  radius 
drawn  to  the  jmmt  tvhere  the  normal  intersects  the  parabola 
and  a  line  through  the  same  point  parallel  to  the  axis  of  the 
jyarahola. 

If,  in  Fig.  79,  PiN  is  the  normal  to  the  parabola  at  Pi  and 
KKi  is  drawn  through  Pj  parallel  to  OX,  we  are  to  prove 
^  FPiN  =:Z  NPiKi,  or  that  PiN  bisects  the  angle  FPiKi. 

Now  ZKPiT  =  ZFTPi,  since  they  are  the  alternate- 
interior  angles  of  two  parallel  lines  cut  by  a  secant ;  and  we 
have  just  proved  /_  FTPj  =  Z.  FP^T. 

Therefore  Z  KPiT  =  Z  FPiT,  and  TPj  bisects  the  angle 
KPiF. 

It  follows  that  PiN  bisects  the  angle  FPjKi,  for  a  line, 
through  the  point  of  intersection  of  two  lines,  perpendicular 


THE   PARABOLA. 


181 


to  the  bisector  of   one  pair  of   the  vertical   angles   formed 
by  the  two  lines  bisects  the  other  pair. 

84.     Perpendicular  Tangents  meet  on  the  Directrix. 


Fig.  80. 

As  y^  =  4px  is  the  equation  of  the  parabola,  the  equation  of 
its  directrix  is  x  =  —  p.     Let  the  equation  of  any  tangent  be 

y  =  mx  +  £-  (1) 

Then,  if  (2)  is  the  tangent  perpendicular  to  (1),  its  equation 

n  1 

will  be  y  =  m'x  +  — „   where  m'  = , 


m' 


or  y  = pm.  (2) 

If  tangents  (1)  and  (2)  meet  at  (xj,  y,),  we  find,  by  solving 
(1)  and  (2)  simultaneously,  that  Xi  =  —  p  and  yi  =  -^  —  pm, 
values  which  satisfy  the  equation  of  the  directrix. 


182 


PLANE   ANALYTIC    GEOMETRY. 


As  (1)  and  (2)  are  any  pair  of  perpendicular  tangents,  it 
follows  that  any  two  perpendicular  tangents  meet  on  the 
directrix,  and  the  directrix  is  the  locus  of  the  point  of  inter- 
section of  perpendicular  tangents. 

85.     Perpendicular  to  Focal  Chord. 

The  perpendicular  to  any  focal  chord  at  the  focus  meets  the 
tangent  at  the  extremity  of  the  focal  chord  on  the  directrix. 


Fig.  81. 


f  =  Apx  (1) 

is  the  parabola, 

x  =  -p  (2) 

is  its  directrix,  and  F  (p,  0)  is  its  focus. 

If  Pi(xi,  Yi)  is  any  point  of  the  parabola,  the  equation  of 
the  tangent  through  Pj  will  be,  by  (1),  §  76, 

y,y  =  2p(x  +  x0.  '  (3) 


THE    PARABOLA.  183 

By  [14],  §  34,  the  equation  of  FPj  will  be 

y^U2-'-:ZJh  ovy-y,  =  -^^(x-xO.  (4) 


yi  — 0     xj  — p 

Since  FR  is  perpendicular  to  FPj  its  equation  will  be,  by 
[18],  §  36, 

y-0  =  P^(x-p).  (5) 

yi 

Solving  (3)  and  (5)  simultaneously,  we  find  the  coordinates 
of  R  to  be  —  p  and  '^-^-^ — ^,  so  that   R   is  a  point  of  the 

yi 

directrix,  as  we  set  out  to  prove. 

This  theorem  gives  us  an  easy  method  of  constructing  the 
tangent  to  the  parabola  when  the  point  of  contact  is  given. 
For  we  have  only  to  draw  the  focal  chord  to  the  point,  then 
draw  through  the  focus  a  line  perpendicular  to  this  focal 
chord,  and  from  the  point  where  this  perpendicular  inter- 
sects the  directrix  draw  a  line  to  the  given  point  of  contact. 
This  last  line  will  be  the  required  tangent. 

86.     Perpendicular  from  Focus  to  Tangent. 

The  perpendicular  from  the  focus  to  any  tangent  ivill  meet  it 
on  the  tangent  at  the  vertex  of  the  parabola,  and  will  meet  the 
directrix  at  the  point  where  the  latter  is  intersected  by  a  line 
from  the  point  of  contact  parallel  to  the  axis  of  the  parabola. 

y2  =  4px  (1) 

is  the  parabola, 

x  =  -p  (2) 

is  its  directrix,  and  F  (p,  0)  is  the  focus. 

The  equation  of  the  tangent  at  the  vertex,  which  is  the 
point  (0,  0),  will  be 

0y  =  2p(x  +  0)  or  x  =  0.'  (3) 


184 


PLANE    ANALYTIC    GEOMETRY. 


Fig.  82. 

If  Pi(xi,  yi)  is  any  point  of  the  parabola,  the  equation  of 
the  tangent  through  P^  will  be 

yiy  =  2p(x  +  xO.  (4) 

Since  FRS  is  perpendicular  to  the  tangent,  its  equation  will 
be 


^  =  -2p^'^-P> 


(5) 


If  R  is  the  point  of  intersection  of  (4)  and  (5),  we  find  R  to 
be  the  point  (  0,  ^  )  and  hence  to  be  a  point  of  (3),  as  was  to 
be  proved. 

If  S  is  the  point  of  intersection  of  (5)  and  (2),  we  find  S  to 
be  the  point  ( —  p,  yi).  Therefore,  S  is  at  the  same  distance 
from  OX  as  is  P^,  and  hence  is  on  the  line  through  Pj  parallel 
to  OX. 


THE   PARABOLA. 


185 


It  is  readily  showu  that  tlie  distances  RS  and  RF  are  both 


V. 


yi 


equal  to  \/p'^  +  ^,  so  the  tangent  is  perpendicular  to  FS  at 

its  middle  point.  And  here  we  find  a  method  of  constructing 
the  tangents  to  a  parabola,  passing  through  a  point  not  on  the 
parabola. 

Y 


Fig.  83. 


Let  Pj  be  the  point  from  which  the  tangents  are  to  be 
drawn,  and  let  F  be  the  focus  of  the  parabola.  With  P^  as  a 
centre  and  a  radius  equal  to  PiF  describe  a  circle,  denoting 
by  $2  and  S3  the  points  at  which  it  meets  the  directrix. 
From  $2  and  S3  draw  lines  parallel  to  OX,  which  will  intersect 
the  parabola  at  P2  and  P3  respectively. 

Then  P1P2  is  one  of  the  required  tangents.  For  P1S2  =  PiF, 
being  radii  of  the  same  circle;  and  S2P2^FP2,  since  the 
eccentricity  of  a  parabola  is  unity.  Therefore,  P1P2  is  per- 
pendicular to  FS2  at  its  middle  point  and  hence  coincides  witli 
the  tangent  at  P2.  Similarly  it  can  be  proved  that  PiPj  is  a 
tangent  to  the  ])arabola. 


186 


PLANE    ANALYTIC    GEOMETRY. 


Note.  If  Pi  is  so  situated  that  tlie  circle  with  radius  PiF  does  not 
cut  the  directrix,  the  above  construction  fails.  But  it  is  evident  that 
this  case  can  occur  only  when  Pi  is  a  point  within  the  parabola,  and 
from  such  a  point  no  tangent  can  be  drawn  to  the  parabola. 

87.     Diameter. 

The  definition  of  a  diameter  which  was  given  in  §  61  for 
the  circle  is  the  definition  of  the  diameter  for  all  the  conies. 
Hence  the  diameter  of  the  'parabola  is  the  locus  of  the  middle 
jyoints  of  a  system  of  parallel  chords. 


Let 


y2  =  4px 


(1) 


be   the   parabola,   and  let   m   be  the  common    slope   of  the 
parallel  chords.     Then 

y  =  mx  +  b  (2) 

will  represent  one  of  the  chords,  as   PiPo,  which  intersects 
the    parabola   at   the   points    Pi(xi,   yi)    and    P2(x2,   ys)-      If 


THE    PARABOLA.  187 

PsC^s;  ys)  is  the  middle  point  of  this  chord, 

Substituting  for  x  from  (2)  into  (1),  we  have 

^  '    y     m 

/-iPy+ie-^=0,  (4) 

the  roots  of  which  equation  are  yi  and  yj. 

.••yi  +  y.  =  ^,  (5) 

whence,  substituting  in  (3),  we  have 

y.  =  ?^.  (6) 

As  this  is  an  expression  for  a  coordinate  of  the  middle 
point  of  one  of  the  chords,  independent  of  b  and  therefore 
the  same  for  all  chords, 

y  =  —  [36] 

is  an  equation  satisfied  by  the  coordinates  of  the  middle  point 
of  every  one  of  the  system  of  chords,  i.e.,  by  the  coordinates 
of  every  point  of  the  diameter,  and  is  accordingly  the  equa- 
tion of  the  diameter. 

Note.  The  work  here  differs  from  that  of  §  61,  in  that  ys  alone  is 
found,  while  there  we  found  xj  and  ys  both.  The  aim  in  both  cases  has 
been  the  same, — to  find  an  equation  in  xs  and  ys  which  shall  be  inde- 
pendent of  b,  and  hence  be  true  for  the  middle  point  of  every  chord  of 
the  system. 

In  the  case  of  the  circle  the  values  of  both  xs  and  ys  involved  b,  so 
that  the  only  way  in  which  we  could  obtain  the  desired  equation  was  by 
eliminating  b  from  the  equations  containing  xs  and  ys.  In  the  case  of 
the  parabola  ys  does  not  involve  b,  so  that  we  have  immediately  an 
equation  of  the  right  form,  and  hence  are  not  obliged  to  find  Xs. 


188  PLANE   ANALYTIC    GEOMETRY. 

It  is  to  be  noted  that  this  diameter  is  parallel  to  the  axis 
of  the  parabola  and  hence  that  all  diameters  of  the  parabola 
are  parallel. 

We  can  now  state  the  theorems  of  §§  83,  86,  in  much  better 
forms,  as  follows  : 

The  tangent  makes  equal  angles  with  the  focal  chord  and 
the  diameter  drawn  through  the  point  of  contact. 

The  normal  makes  equal  angles  with  the  focal  chord  and 
the  diameter  drawn  through  the  point  where  the  normal  inter- 
sects the  parabola. 

The  perpendicular  from  the  focus  to  any  tangent  will  meet 
it  on  the  tangent  at  the  vertex  of  the  parabola,  and  will  meet 
the  directrix  at  the  point  where  the  directrix  is  cut  by  the 
diameter  drawn  through  the  point  of  contact. 

Ex.  1.     Find  the  diameter  of  the  parabola  y-  =  —  7x,  bisecting  chords 

parallel  to  the  line  x  —  y  +  2  =  0. 

For  the  chords  m  =  1,  and  p  =  —  |. 

2(— 1) 
.-.  the  diameter  is  y  =         •*  ,  or  2y  +  7  =  0. 

Ex.  2.  A  diameter  of  the  parabola  y-  =  6x  passes  through  the  point 
(2,  —  1).  What  is  its  equation,  and  what  is  the  slope  of  the  chords 
bisected  by  it  ? 

3 
If  m  is  the  slope  of  the  chords  bisected,  the  diameter  is  y  =  —  •     But 

m 

(2,  —  1)  is  a  point  of  the  diameter.  \, 

3  3 

.•.  —  1  =  —  and  m  =  —  3,  and  the  diameter  is  y  =  — -,  or  y  =  —  1. 
m  —  3 

We  could  have  written  down  this  last  immediately,  for  the  diameter 
is  parallel  to  OX,  so  that,  if  one  of  its  points  is  distant  —  1  from  OX,  all 
its  points  are  distant  —  1  from  OX,  and  its  equation  is  y  =  —  1. 

88.     Properties  of  the  Diameter. 

The  tangent  at  the  extremitu  of  a  diameter  is  imrallel  to  the 
chords  bisected  by  the  diameter. 

2p 
If  the  equation  of  the  diameter  is  y  =:— ^,  it  will  intersect 

m 


THE    PARABOLA.  189 

the  parabola  at  the  point  (2'         )•     The  tangent  to  the 

parabola  at  this  point  is  —  y  =  2p  (  x  +     2  )  or  y  =  mx  +  ^, 

the  slope  of  which  is  m,  or  the  slope  of  the  chords  bisected 

by  the  diameter. 

The,  polar  of  any  point  of  a  diameter  is  pjarallel  to  the  chords 

bisected  by  the  diameter. 

"p\  -P 

x,,  —  )  is  any  point  of  the  diameter  y  =  --. 
my  m 

Its  polar  will  be 

2p 

— !- y  =  2p  (x -}- Xi)  or  y  =  mx -|- inx, , 
m 

which  line  has  its  slope  equal  to  m,  the  slope  of  the  chords 
bisected  by  the  diameter. 

It  follows  that  the  pole  of  every  chord  of  the  parabola  is 
on  the  diameter  wliich  bisects  the  chord.  As  a  chord  of  con- 
tact is  a  special  case  of  a  polar,  the  corresponding  tangents 
meeting  at  the  pole,  the  tangents  at  the  extremities  of  any  chord 
meet  on  the  diameter  ^vhich  bisects  the  chord. 


89.     Parabola  with  Focus  at  the  Origin. 

This  form  of  equation  of  the  parabola  is  readily  found  by 
transforming  the  equation  y^  =  4px  to  F  (p,  0)  as  a  new  origin, 
the  directions  of  the  axes  not  being  changed.  The  formulas 
of  transformation  are  x  =  x'  +  p  and  y  =  y',  so  tliat  the  trans- 
formed equation  is,  after  the  primes  are  dropped, 

y2  =  4px+4pl  (1) 

It  is  evident  that  y^  =  —  4piX-}--lpi^  is  the  equation  of  a 
parabola  having  its  focus  at  the  origin,  but  extending  indefi- 
nitely toward  the  left,  as  it  is  the  form  (1)  assumes  if  the 
sign  of  p  is  changed,  p  and  pi  being  positive. 


190 


PLANE   ANALYTIC   GEOMETRY. 


Fig.  85. 


Any  two  parabolas  having  the  same  focus  are  said  to  be 
confocal,  and  the  above  is  a  special  case  of  confocal  parabolas. 
Moreover,  if  we  define  the  angle  between  two  intersecting 
curves  as  the  angle  between  their  respective  tangents  at  their 
point  of  intersection,  two  such  confocal  parabolas  as  above 
will  intersect  at  right  angles. 

Solving  y^  =  4px  +  4p'^  and  y-  =  — 4piX  +  4pi2,  we  find  these 
parabolas  meet  at  the  points 

(pi  -  p,  2  Vpip)  and  (pi  —  p,  -  2  Vpip). 

The  respective  tangents  at  the  point  (pi  —  p,  2  Vpip)  are 

by  [34],  §  76,        

2  Vpip  y  =  2p  (x  +  pi  -  p)  +  4p2, 
and  2  Vpipy  =  — 2pi  (x  + pi  — p)  +  4pi2, 


THE    PARABOLA. 


191 


are 


or  Vpipy  =  px  +  ppi+ p2, 

and  Vpipy  =  — pix  +  pi-+ pip. 

Their  slopes  are  respectively -i/—  and  — \/     5  so  they 

perpendicular  to  each  other,  and  hence  the  parabolas  cross  at 
right  angles. 

In  like  manner  the  parabolas  may  be  shown  to  cross  at 
right  angles  at  the  other  point  of  intersection, 

(Pi-  p, -2Vpip). 


90.  Parabola  Referred  to  a  Diameter  and  the  Correspond- 
ing Tangent  as  Axes. 

In  this  article  we  will  find  the  equation  of  the  parabola 
"with  respect  to  a  diameter  as  the  axis  of  x  and  the  tangent  at 
the  extremity  of  the  diameter  as  the  axis  of  y. 

Y' 


Fig.  86. 


192  PLANE    ANALYTIC    GEOMETRY. 

2p 
If  the  equation  of  the  diameter  O'X'  is  y  =  -^,  the  equation 

of  the  tangent  O'Y'  will  be  y  —  mx  +—  and  the  coordinates 

of  0' will  be  (^^o,— \ 
V  m''     my 

If  we  take  first  a  set  of  axes  O'X'  and  O'Y"  parallel  to  the 

original  axes,  we  shall  transform  the  equation  y^  :=  4px  by  the 

P  2p 

formulas  x  =  -^  +  x"  and  y  =  -^  +  y", 

obtaining  the  equation 

■'  m 

To  transform  this  last  equation  from  axes  O'X'  and  O'Y"  to 
O'X'  and  O'Y'  the  formulas  will  be 

y'  y'm 

for  in  formulas  [21],  §  45,  ^  =  0  and  6'  =  tau"^  m. 
The  result  of  this  substitution  will  be 

By[l],§3,       FO'  =  P-(i±^\ 

Therefore  if  we  denote  FO'  by  p',  our  equation  becomes, 
after  dropping  the  primes, 

y^  =  4p'x.  [37] 

It  is  to  be  noted  that  y'^  =  4px  ahvays  represents  a  parabola, 
the  X  axis  being  a  diameter,  the  y  axis  a  tangent,  and  the 
distance  of  the  focus  from  the  origin  being  one  fourth  the 
coefficient  of  x. 


THE    PARABOLA.  193 

91.     Parabola  in  Polar  Coordinates. 

For  the  polar  equation  of  the  parabola  we  will  take  the  axis 
of  the  parabola  as  the  initial  line  and  the  focus  as  the  origin. 
In  Fig.  79  draw  PiM  perpendicular  to  OX  ;  then 

FPi  =  p  +  Xi  =  p  +  OM  =  p  +  OF+  FM=2p+  FM. 

But  in  polar  coordinates  FPj  =  r  and  FM  =  r  cos  $,  and  we 

have  r  =  2p  +  r  cos  6, 

whence  we  derive  as  our  required  equation  of  the  parabola 

„_       2p 


1  —  cos  6 


EXAMPLES. 


[38] 


1.  Find  the  length  of  the  chord  of  the  parabola  y^  +  2x=0 
cut  from  the  line  4x  —  3y  -|-  2  =  0. 

2.  If  the  focal  distance  of  a  point  on  a  parabola  y^  =  4px 
is  n  times  the  latus  rectum  of  the  parabola,  what  are  its  coor- 
dinates ? 

3.  Two  straight  lines  are  drawn  throvigh  the  vertex  of  a 
parabola  at  right  angles  to  each  other  and  meeting  the  curve 
at  P  and  Q.  Show  that  the  line  PQ  cuts  the  axis  of  the 
parabola  in  a  fixed  point. 

4-  Yi,  y25  and  yg  are  the  ordinates  of  three  points  of  the 
parabola  y^  =  4px.  Prove  that  the  area  of  the  triangle 
formed  by  joining  them  in  succession  is 

grCyi-ys)  (y^-ya)  (ys-yO- 

5.  Taking  the  definitions  of  subtangent  and  subnormal 
given  for  the  circle,  prove  : 

(a)  That  the  subtangent  of  any  point  of  a  parabola  is 
twice  the  projection  on  the  axis  of  the  parabola  of  the  line 
joining  the  point  to  the  vertex. 


194  PLANE    ANALYTIC    GEOMETRY. 

(b)  That  the  subnormal  always  equals  the  distance  from 
the  focus  to  the  directrix. 

6.  Find  the  equations  of  the  tangent  and  the  normal  to 
the  parabola  y^  =  9x  which  pass  through  (4,  6). 

7.  Find  the  equation  of  a  tangent  to  the  parabola  y^^  6x 
which  passes  through  (2,  4). 

v>^        8.  Find  the  equation  of  a  tangent  to  the  parabola  y^  =  9x 
/^      perpendicular  to  the  line  2x  +  y  —  2  =  0. 

9.  A  tangent  to  the  parabola  y^  ^  7x  makes  an  angle  of 

60°  with  the  axis  of  the  parabola.     What  are  the  coordinates 

of  its  point  of  contact  ? 

10.  Find  a  normal  to  the  parabola  y^  =  4x  which  shall  pass 
through  (2,  —  8). 

11.  Find  a  normal  to  the  parabola  y'-=;9x  parallel  to  the 
line  X  +  y  —  1  =  0. 

12.  What  are  the  slopes  of  tlie  tangents  drawn  to  the 
parabola  y^  =  x  from  the  point  ( —  1,2)? 

13.  Find  the  tangents  and  normals  at  tlie  extremities  of 
the  latus  rectum  of  the  parabola  y^  =  4px. 

14.  If  perpendiculars  be  let  fall  on  any  tangent  to  a  para- 
bola from  two  given  points  on  the  axis  which  are  equidistant 
from  the  focus,  the  difference  of  their  squares  is  constant. 

15.  If  the  chord  PQ  is  a  normal  to  the  parabola  at  P,  and 
the  tangents  at  P  and  Q  meet  at  T,  show  that  PT  is  bisected 
by  the  directrix. 

16.  Find  the  coordinates  of  the  point  where  the  normal  at 
(xi,  yi)  again  meets  the  parabola  y^  =  4px,  and  the  length  of 
the  intercepted  line. 

17.  M  P  is  the  ordinate  of  a  point  P  of  the  parabola  y^=:  4px. 
A  straight  line  is  drawn  parallel  to  the  axis  and  bisecting  M  P, 
which  cuts  the  curve  at  Q.  If  MQ  cuts  the  tangent  at  the 
vertex  A  at  T,  show  that  AT  =  f  M  P. 

18.  Show  that  the  locus  of  the  point  of  intersection  of  two 
tangents  to  a  parabola,  the  ordinates  of  the  points  of  contact 
of  which  are  in  a  constant  ratio,  is  a  parabola. 


THE    PARABOLA.  195 

19.  Show  that  perpendicular  normals  to  the  parabola 
y^  =  4px  intersect  on  the  curve  y^  =  px  —  3p^. 

20.  Prove  that  the  locus  of  the  point  of  intersection  of  two 
tangents  to  a  parabola  is  a  straight  line  if  the  product  of  their 
slopes  is  constant. 

21.  Prove  that  the  ordinate  of  the  point  of  intersection  of 
two  tangents  to  a  parabola  is  the  arithmetical  mean  between 
the  ordinates  of  the  points  of  contact  of  the  tangents. 

22.  If  P,  Q,  and  R  are  three  points  on  a  parabola,  the  ordi- 
nates of  which  are  in  geometrical  progression,  show  that  the 
tangents  at  P  and  R  will  meet  on  the  ordinate  of  Q. 

23.  Prove  that  the  area  of  the  triangle  formed  by  the  axis 
of  a  parabola  and  the  tangent  and  the  normal  of  any  point,  is 
the  same  as  the  area  of  the  rectangle  forined  by  the  directrix, 
the  axis  of  the  parabola,  and  the  two  lines  through  the  point 
respectively  i)arallel  to  the  directrix  and  the  axis. 

24.  Find  the  equation  of  the  line  through  the  vertex  and 
the  upper  extremity  of  the  latus  rectum  of  the  parabola 

y^  =  4px. 

25.  How  far  are  the  tangents  at  the  extremities  of  the 
latus  rectum  of  the  parabola  y^  +  3x  :=  0  from  the  vertex  of 
the  parabola  ? 

26.  The  tangent  at  any  point  of  the  parabola  y^  =  4px  will 
meet  the  directrix  and  the  latus  rectum  produced  in  two 
points  equidistant  from  the  focus. 

27.  Show  that  the  normal  at  one  extremity  of  the  latus 
rectum  of  a  parabola  is  parallel  to  the  tangent  at  the  other 
extremity  of  the  latus  rectum. 

28.  The  normal  at  L,  the  upper  end  of  the  latus  rectum, 
meets  the  parabola  again  at  P.  Show  that  the  diameter  on 
which  the  tangents  at  L  and  P  intersect  passes  through  the 
other  extremity  of  the  latus  rectiim. 

29.  Show  that  the  tangents  at  tlie  ends  of  the  latus  rectum 
are  twice  as  far  from  the  focus  as  from  the  vertex. 


196  PLANE    ANALYTIC    GEOMETRY. 

30.  Find  the  equation  of  the  circle  through  the  vertex, 
the  focus,  and  the  upper  end  of  the  latus  rectum  of  the  para- 
bola y2+ 12x  =:0. 

31.  Find  the  equation  of  the  circle  through  the  vertex  and 
the  ends  of  the  latus  rectum  of  the  parabola  y^  =  4px. 

32.  Find  the  length  of  the  perpendicular  from  the  focus  of 
the  parabola  y'^  =  4px  to  the  tangent  at  any  point  (xi,  yi)  in 
terms  of  Xi  and  p. 

33.  Prove  that  the  line  joining  the  focus  to  the  intersection 
of  two  tangents  bisects  the  angle  which  their  points  of  con- 
tact subtend  at  the  focus. 

34.  If  tangents  are  drawn  at  the  ends  of  two  focal  radii, 
show  that  the  angle  between  the  radii  is  double  the  angle 
between  the  tangents. 

35.  Find  the  equation  of  the  circle  circumscribing  the 
triangle  formed  by  the  axis  of  the  parabola  y2  =  3x,  the 
tangent  at  the  point  (i-,  1),  and  the  focal  radius  drawn  to 
that  point. 

36.  Show  that  any  circle  described  on  a  focal  chord  as 
diameter  is  tangent  to  the  directrix  of  the  parabola. 

37.  Show  that  the  circle  described  on  any  focal  radius  as 
diameter  is  tangent  to  the  tangent  at  the  vertex  of  the 
parabola. 

38.  A  circle  has  its  centre  at  the  vertex  A  of  a  parabola 
y2=z4px,  and  the  diameter  of  the  circle  is  3  AF.  Show  that 
the  common  chord  bisects  AF. 

39.  A  diameter  of  the  parabola  3y^  =  5x  bisects  chords  per- 
pendicular to  the  line  x  +  3y  +  1  ==  0.     What  is  its  equation  ? 

40.  A  diameter  of  the  parabola  y^  +  7x  =  0  passes  through 
the  point  (2,  1).  Find  its  equation  and  the  slope  of  the 
chords  which  it  bisects. 

41.  y  +  3  is  a  diameter  of  the  parabola  2y^  +  3x  =  0.  What 
is  the  slope  of  the  chords  which  it  bisects  ? 

42.  (2,  —  5)  is  the  middle  point  of  a  chord  of  the  parabola 
y2  —  3x  =  0.     What  is  the  equation  of  the  chord  ? 


THE   PARABOLA.  197 

43.  Find  the  diameter  of  the  parabola  Sy'-'  +  5x  =  0  which 
bisects  chords  parallel  to  the  polar  of  (—  1,  —  5). 

44.  Find  the  equation  of  the  tangent  to  the  parabola 
y2  —  5x  =  0  parallel  to  the  chords  bisected  by  the  diameter 
y+4=:0. 

45.  Show  that  the  tangents  at  the  extremities  of  any  chord 
of  a  parabola  meet  on  the  diameter  which  bisects  that  chord. 

46.  If  tangents  are  drawn  at  the  extremities  of  any  focal 
chord  of  a  parabola,  show  :  (1)  that  the  tangents  will  inter- 
sect on  the  directrix  ;  (2)  that  the  tangents  will  be  perpendic- 
ular to  each  other  ;  (3)  that  the  straight  line  drawn  from  the 
point  of  intersection  of  the  tangents  to  the  focus  will  be  per- 
pendicular to  the  focal  chord. 

47.  Prove  that  the  portion  of  the  axis  included  between  the 
polars  of  two  points  eqiials  the  projection  on  the  axis  of  the 
line  joining  the  points. 

48.  Find  the  locus  of  the  feet  of  the  perpendiculars  from 
the  focus  to  the  normals  of  the  parabola  y'^  =  4px. 

49.  P  is  any  point  of  a  parabola,  A  the  vertex,  and  through 
A  a  straight  line  is  drawn  perpendicular  to  the  tangent  at  P. 
Find  the  locus  of  the  point  of  intersection  of  this  line  with 
the  diameter  through  P,  and  also  the  locus  of  the  point  of 
intersection  of  this  line  with  the  ordinate  through  P. 

50.  Two  parabolas  have  the  same  axis,  and  tangents  are 
drawn  from  points  on  the  first  to  the  second.  Prove  that  the 
middle  points  of  the  chords  of  contact  with  the  second  lie  on 
a  parabola. 

51.  If  the  tangent  to  the  parabola  y^^4px  meets  the  axis 
at  T  and  the  tangent  at  the  vertex  A  at  B,  and  the  rectangle 
TABQ  be  completed,  show  that  the  locus  of  Q  is  the  parabola 
y^  -[-  px  =  0. 

52.  A  point  moves  so  that  its  shortest  distance  from  a 
fixed  circle  is  equal  to  its  distance  from  a  fixed  diameter  of 
that  circle.  Find  its  locus.  (Take  tlie  diameter  of  the  circle 
as  the  axis  of  x  and  the  centre  of  the  circle  at  0.) 


198  PLANE   ANALYTIC    GEOMETRY. 

53.  If  a  straight  line  be  drawn  from  the  origin  to  any  point 
Q  of  the  line  y  =  a,  and  if  a  point  P  be  taken  on  this  line  such 
that  its  ordinate  is  equal  to  the  abscissa  of  Q,  show  that  the 
locus  of  P  is  a  parabola. 

54.  Two  equal  parabolas  have  their  axes  parallel  and  a 
common  tangent  at  their  vertices,  and  straight  lines  are  drawn 
parallel  to  the  axes.  Show  that  the  locus  of  the  middle  points 
of  the  parts  of  the  lines  intercepted  between  the  curves  is  an 
equal  parabola. 

55.  Find  the  locus  of  the  centre  of  a  circle  which  shall 
always  be  tangent  to  a  given  circle  and  a  given  straight  line. 

56.  Prove  that  the  locus  of  the  centre  of  a  circle  which 

always  passes  through  a  given  point  and  is  tangent  to  a  given 

straight  line  is  a  parabola. 

p 

57.  If,  in  the  triangle  ABC,  tan  A  tan  iT  =  2,  and  AB  is 

A 

fixed,  show  that  the  locus  of  C  is  a  parabola,  the  vertex  of 
which  is  A  and  the  focus  B.  (Take  A  as  the  origin  and  AB  as 
the  axis  of  x.) 

58.  The  base  of  a  triangle  is  2b  and  the  sum  of  the  tan- 
gents of  the  angles  at  the  base  is  s.  Find  the  locus  of  the 
vertex.  (Take  the  base  of  the  triangle  as  the  axis  of  x  and 
the  middle  point  of  the  base  as  origin.) 

59.  Find  the  eqiiation  of  the  parabola  referred  to  the  tan- 
gents at  the  ends  of  the  latus  rectum  as  coordinate  axes. 


CHAPTER   IX. 


THE    ELLIPSE  :  —  +  i-  =  1. 


92.     We  have  seen,  in  §  68,  chap.  VI,  that  the  equation 
of  the  ellipse  takes  its  simplest  fonu 

when  the  centre  of  the  curve  is  at  the  origin  of  coordinates 

and  the  major  axis  of  the  curve  coincides  with  the  axis  of  x. 

The  foci  of  the  ellipse  are  then  at  the  points  (±  ae,  0),  and  the 

.    n      n-          •                          a       ,                 Va^  -  b^ 
equations  or  the  directrices  are  x  =  ±  - ,  where  e  = 

We  have  further  seen,  §  78,  chap.  VII,  that  tlie  polar  of  the 

•  1  1  •  •     ^iX   .   y,y        .    ^, 

point  (xj,  Yi)  with  respect  to  this  curve  is  — - -\-  \J,  ="  1,  the 

a"         0" 

polar  coinciding  with  the  tangent,  if  the  point  (xi,  yi)  is  on 

the  curve.     Finally,  if   m  is  the  slope  of  any  tangent,  the 

equation  of  the  tangents   with   that  slope   may   be   written 

y  =  mx  ±  Va^m^  +  bl     [(5),  §  76.]     These  simple  equations 

will  be  used  throughout  this  chapter. 


93.     To  find  the  Foci  when  the  Axes  are  given. 

Let  the  axes  A  A'  and  BB'  (Fig.  87)  be  given  both  in  length 
and  position  ;  it  is  required  to  find  the  foci  F  and  F'. 

By  §68,  CF'  =  ae  =  Va^-b-. 

But  BC  =  b.  Therefore,  from  the  right  triangle  BCF', 
B  F'  =  a.  Similarly,  B  F  =  a.  Hence  we  may  give  the 
simple  rule  : 


200 


PLANE   ANALYTIC   GEOMETKY. 


iVith  either  extremity  of  the  minor  axis  as  a  centre,  and  with 
a  radius  equal  to  the  semi-major  axis,  describe  an  arc  of  a  circle. 
The  arc  will  cut  the  major  axis  in  the  foci. 

94.     Focal  Distances. 

The  distances  of  a  point  on  an  ellipse  from  the  foci  are 
called  its  focal  distances,  and  the  lines  joining  the  point  to 
the  foci  are  called  the  focal  radii.  The  focal  distances  are 
easily  expressed  in  terms  of  the  abscissa  of  the  point  and  the 
constants  of  the  ellipse.  For,  consider  the  ellipse  of  Fig.  88, 
in  which  the  directrices  DM  and  D'M'  have  been  drawn. 


Fig.  88. 


THE    ELLIPSE.  201 

By  the  definition  of  the  ellipse, 

But  MPi=  DN  =  DC+CN, 

e 
Therefore,  FPi  =  a  +  exi. 

Similarly,  F'Pi  =  e.  PjM', 

=  e  (CD'-CN), 

=  ^(e-'^' 
=  a  —  exi. 

95.     Sum  of  Focal  Distances. 

From  the  results  of  the  previous  article  follows  at  once 
a  very  simple  property  of  the  ellipse.  For,  if  we  add  our 
two  results,  we  obtain 

FPi+  F'Pi  =  2a. 

That  is  :  tJie  sum  of  the  focal  distances  of  any  jwhit  on  an 
ellipse  is  constant  and  equal  to  the  major  axis. 

Moreover,  it  is  evident  that  this  property  is  not  true  for 
any  point  not  on  the  ellipse.  This  follows  from  the  theorem 
of  Geometry,  that  the  sum  of  two  lines  drawn  from  the 
extremities  of  a  straiglit  line  is  less  than  the  sum  of  two 
lines  similarly  drawn,  but  included  by  them.  Hence  we 
may  say  : 

The  ellipse  is  the  locus  of  a  point  the  sum  of  the  distances  of 
which  from  two  fixed  points  is  constant. 

This  is  sometimes  given  as  the  definition  of  the  ellipse,  and 
the  equation  and  properties  of  the  ellipse  deduced  from  it. 
Such  a  method  of  derivation,  however,  fails  to  bring  out  the 
relation  of  the  ellipse  to  the  other  conic  sections  as  shown  by 
the  definition  of  chapter  VI. 


202 


PLANE   ANALYTIC    GEOMETRY. 


96.     Constructions. 

The  last  article  forms  the  basis  of  two  methods  of  con- 
structing an  ellipse  when  the  two  axes  are  given,  or  when 
the  foci  and  major  axis  are  given. 

(1)  By  compasses.  Find  the  foci,  if  they  are  not  already 
given.  Then  take  a  line  equal  to  the  major  axis  and  with 
compasses  divide  it  into  two  parts,  r  and  r'.  With  one  focus 
as  centre  describe  an  arc  of  a  circle  of  radius  r,  and  with  the 
other  focus  as  centre  an  arc  of  radius  r'.  The  intersections  of 
the  two  arcs  are  points  on  the  ellipse.  Repeated  applications 
of  this  process  give  as  many  points  as  are  wished. 

(2)  By  a  mechanical  device.  Find  the  foci  by  §  93,  if  they 
are  not  already  given.  Fix  a  pin  at  each  focus  and  attach  to 
each  pin  one  end  of  a  thread  of  length  2a.  Place  the  point 
of  a  pencil  in  the  loop  of  the  thread  and  move  it  so  as  to 
keep  the  thread  always  taut.  The  pencil  then  traces  an 
ellipse. 


97.     Angle  between  Tangent  and  Focal  Radii. 

The  tangent  to  an  ellq^se  snakes  equal  angles  with  the  focal 
radii  of  the  point  of  contact. 

Y 


Fig.  89. 


THE    ELLIPSE.  203 

Let  TT'  be  a  tangent,  Pi(xi,  yi)  the  point  of  tangency,  and 
FPi  and  F'Pi  the  focal  radii  of  Pi.     To  prove 
ZFPiT'^ZF'PiT. 
Draw  the  normal,  PiR.     By  (2),  §  77,  its  equation  is 

Call  CR  its  intercept  on  the  axis  of  x  ;  then,  placing  y  ^=  0, 

we  find 

ro       a-— b- 

CK  = :; —  Xi  =  e-Xi. 

a" 

Hence  FR  =  FC  +  CR, 

=  ae  +  e"Xi  ; 

and  RF'=CF'-CR, 


ae  —  e-x, 


Therefore, 


FR        ae  +  e^xi       a  +  exi       FPj 


RF'       ae  — e%       a  —  exj      F'Pi 

Hence  Z  RPiF  =  Z  RPiF' ; 

for,  if  a  line  through  the  vertex  of  a  triangle  divides  the 
opposite  side  into  segments  proportional  to  the  adjacent 
sides,  it  bisects  the  angle  at  the  vertex. 

Therefore,  finally, 

ZFPiT'  =  ZF'PiT. 

It  may  be  noticed,  in  passing,  that  the  value  of  CR  shows 
tluit  the  normal  does  not  pass  through  the  centre  of  the 
ellipse,  but  cuts  the  major  axis  between  the  centre  and  the 
foot  of  the  ordinate  from  the  given  point.  It  is,  in  fact,  a 
distinct  characteristic  of  the  circle  that  the  normal  always 
passes  through  the  centre. 

98.  Point  of  Intersection  of  a  Pair  of  Perpendicular 
Tangents. 

Let  Pi(xi,  yi)  be  the  point  of  intersection  of  a  pair  of  per- 
pendicular tangents  to  the  ellipse  ;  it  is  required  to  find  the 


204 


PLANE   ANALYTIC   GEOMETRY. 


Fig.  90. 


locus  of  Pp     The  equation  of  any  tangent  to  the  ellipse  may 
be  written  in  the  form 


y  =  mx  it  Va^m^  +  b'^. 
If  the  tangent  passes  through  P^,  we  shall  have 
yi  =  mxi  zt  Va^m^  -\-  b^, 
which  is  an  equation  from  which  the  value  of  m  may  be  com- 
puted in  terms  of  Xi  and  yi.     In  fact,  the  equation  reduces  to 
the  quadratic  in  m, 

(a^  —  xf)  m-  +  2xiyim  +  b-  —  y^^  =  0, 

the  two  roots,  trii  and  nria,  of  which  correspond  to  the  two 
tangents  through  P^.  Now,  by  the  theory  of  quadratic 
equations, 

2xiyi 


iTii  +  rrio  =  — 


mim2 


a^ —  Xi 

b^  —  Yi' 
a^  —  x,^* 


THE    ELLIPSE.  205 

But,  by  hypothesis,  the  two  tangents  are  perpendicular. 
Hence  mirng  =:  —  1,  and  therefoi-e, 

a'^-xr~        ' 
or  Xi^  +  yi"=  a^+  h'\ 

Hence  the  point  Pi  must  lie  upon  the  circle 
x^  +  y-  =  a-  +  b-, 
and  hence  the  theorem  : 

The  locus  of  the  j^oint  of  intersection  of  a  pair  of  2)6vpendicvr 
lar  tangents  to  an  ellipse  is  a  circle  tvith  its  centre  at  the  centre 
of  the  ellipse  and  its  radius  equal  to  the  square  root  of  the  sum 
of  the  squares  of  the  semi-axes  of  the  elUjyse. 

99.     Perpendicular  from  the  Focus  upon  any  Tangent. 


Let  y  =  mx  ±  Va^m='  +  b^  (1) 

be  any  tangent  to  the  ellipse.     Then  the  equation  of  the  per- 
pendicular from  the  right-liand  focus  upon  it  is,  by  [18],  §  36, 

y  =  --(x-ae), 

or  my  -f-  x  =  ae-  (2) 

It  is  required  to  find  the  locus  of  the  point  of  intersection 
of  (1)  and  (2).  If  Xi,  yi  are  the  coordinates  of  the  required 
point,  then  we  have  the  two  equations 


yi  —  mxi  =  ±  Va^m''^  +  b^, 

myi  +  xi  =  ae, 

out  of  which  the  parameter  m   must  be  eliminated.     This 
elimination  may  be  effected  by  squaring  and  adding,  and  then 

^2 b2 

substituting  e^  ^= ^ — .     We  have,  then, 

a 

(1  4-  m-)  Xi"  +  (1  +  m-)  yi'-  =  a''m-  +  b"  +  a^e", 


206 


PLANE   ANALYTIC    GEOMETRY. 


or  (1  +  m^)  X,'  +  (1  +  m^)  y,'  =  (1  +  m")  a', 

whence,  by  dividing  by  (1  +  m^), 

The  same  result  would  be  obtained  by  using  the  left-hand 
focus  (—  ae,  0).     Hence  : 

The  perpendicular  from  either  focus  upon  any  tangent 
intersects  that  tangent  on  the  circle  described  with  the  centre 
of  the  ellipse  as  a  centre  and  the  major  axis  as  a  diameter. 


100.     Equation  of  a  Diameter. 

We  have  already  defined  a  diameter  of  a  conic  as  the  locus 
of  the  middle  points  of  a  system  of  parallel  chords.  To  find 
the  equation  of  a  diameter,  we  proceed  as  in  §  61.  If  nrii  is 
the  slope  of  the  parallel  chords,  then 
y  =  rui  X  +  c 
is  the  equation  of  one  chord,  which  intersects  the  ellipse,  we 
will  say,  in  the  points  Pi(xi,  yi)  and  P2(x2,  yg). 

Y 

P, 


Fig.  91. 


If  X3  and  ya  be  the  coordinates  of  the  middle  point  of  the 
chord,  then 

xi  -f-  X2 


X3 


2 

yi  +  y2 


THE    ELLIPSE.  207 

But  Xi  and  Xo  are  the  roots  of  the  equation 

(b^  +  a-mr)  x2  +  2a-miCx  +  a'  (c-  -  b^)  =  0 ; 
so  that 

x 
and  hence 

Similarly, 

whence,  by  division, 

Xg  a-irii 

This  equation  is  satisfied  by  the  middle  point  of  any  one 
of  the  parallel  chords,  since  it  does  not  involve  c.     Hence, 

y  = :r-  X  [39] 

■^  a-rrii  ■-     -^ 

is  the  equation  of  the  required  diameter.  This  equation 
shows  that  t/ie  diameter  is  a  straight  line  passi7i(jf  through  the 
centre  of  th  e  ellipse. 

In  [39]  iDi  is  the  slope  of  the  chords  which  the  diameter 
bisects.     If  nrig  is  tlie  slope  of  the  diameter  itself,  then 


4-  X   - 

2a^miC 

1      ^^2  - 

b-'  +  a^mi^' 

a-miC 

Xg  —  - 

b- +  a'^rrii^' 

b^c 

ys  — 

b'+a-mi^' 

y3. 

b^ 

or 

b- 

nriimo  = r,. 

a" 

Out  of  this  relation  may  be  deduced  the  important  fact 
that  a  diameter  of  an  ellipse  is  in  general  not  perpendicular 
to  the  chord  it  bisects.  For,  if  it  were,  then  mirrio  would 
equal  —  1. 

This  is  always  true,  if  b  =  a ;  but,  in  that  case,  the  ellipse 
becomes  a  circle,  and  we  are  led  back  to  the  case  discussed  in 
chapter  V. 


208  PLANE    ANALYTIC    GEOMETRY. 

But  if  b  is  unequal  to  a,  rnima  is  not  equal  to  —  1,  and  the 
two  lines  are  not  perpendicular  to  each  other. 

An  exception  occurs  when  the  chords  are  parallel  to  one  of 
the  axes  of  the  ellipse.  In  that  case  mi  =  0  or  co,  and  the 
work  of  finding  the  diameter  cannot  be  carried  out  exactly  as 
above. 

It  is,  however,  easy  to  prove  that  the  diameter  is  then  the 
axis  to  which  the  chords  are  perpendicular.     Hence : 

The  two  axes  of  an  ellipse  are  the  only  diameters  perpen- 
dicular to  the  chords  they  bisect. 

Ex.     Find  a  diameter  of  the  ellipse  ^  +  77  =  1,  bisecting  chords  per- 

25       9 

pendicular  to  the  line  y  +  3x  +  7  =  0. 

The  slope  of  the  given  line  is  —  3,  therefore  the  slope  of  the  parallel 
chords  is  ^.     Substituting  in  the  equation 

b2 


mim2-  — -^, 

we  have 

im2  =  -  i^, 

or 

mg  =  -  ||. 

.  •.  the  diameter  is 

y  =  -  l|x, 

or 

27x  +  25y  =  0. 

101.     Conjugate  Diameters. 

We  have  seen  that,  if  rrii  represents  the  slope  of  a  chord, 
the  slope  m2  of  a  diameter  bisecting  it  is  given  by  the  relation 

a  iDi 

Similarly,  if  nria  is  the  slope  of  a  chord  and  nrix  the  slope  of 

a  bisecting  diameter,  then 

b^ 

mi  = ^ — . 

a^'roo 

Each  of  these  conditions  is  equivalent  to 

b^ 

mirrio  = r,. 

a^ 


THE    ELLIPSE.  209 

Hence  we  may  assert  the  proposition  : 

If  the  slopes  of  two  diameters  are  comiected  by  the  relation 

^1^2  =  -  -2'  [40] 

then  each  diameter  bisects  all  chords  parallel  to  the  other.     Such 
diameters  are  called  conjugate  diameters. 

From  the  condition  [40]  follows  : 

1.  The  two  axes  are  the  only  pair  of  conjugate  diameters 
which  are  perpendicular  to  each  other.  The  proof  is  the 
same  as  in  the  previous  article. 

2.  If  one  of  two  conjugate  diameters  makes  an  acute  angle 
with  the  axis  of  x,  the  other  makes  an  obtuse  angle  with  the 
axis  of  X.     For,  if  mi  is  positive,  m.,  is  negative,  since 

b- 

171^1X12  = 7,. 

a- 

But  a  positive  slope  corresponds  to  an  acute  angle,  and  a 
negative  slope  to  an  obtuse  angle.  Hence  the  upper  portions 
of  conjugate  diameters  always  lie  on  opposite  sides  of  the 
minor  axis  ;  for  example,  CP3  and  CQ  in  Fig.  91. 


102.  The  Tangent  at  the  Extremity  of  any  Diameter  is 
Parallel  to  the  Conjugate  Diameter. 

Let  P1P2  (Fig.  92)  be  any  diameter,  QiQo  its  conjugate 
diameter,  and  PiN  the  tangent  at  Pi.  Then,  if  we  call  the 
coordinates  of  Pi,  (xi,  yi),  the  equation  of  CPi  is 

and  therefore  its  slope 

yi 

mi  =  ^. 

X, 


210 


PLANE    ANALYTIC    GEOMETRY. 


Fig.  92. 


To  find  the  slope  mo  of  CQi,  we  have,  by  substitution  in 
[40],  §  101, 


ll 

m2  =  — 

b^ 

or 

The  tangent  at  Pj  is 

(Da 

__b 
a 

-Xl 

^yi' 

x,x 
3? 

1  yiy. 

=  1, 

and  its  slope, 

therefore 

,  is 

b^x, 

This  is  equal  to  mo.     Hence  the  theorem  is  proved. 


103.     Supplemental  Chords. 

Chords   drawn  from   the   same  point  on  an   ellipse  to  the 
extremities  of  any  diameter  are  called  supplemental  chords. 


THE    ELLIPSE. 


211 


We  shall  prove  that  diameters  2}cvallel  to  a  pair  of  stqyjyle- 
mental  chords  are  conjugate.  To  prove  this  we  need  simply 
to  show  that  the  slopes  of  the  two  chords  satisfy  the  relation 

mim2  =  — 7^- 


Fig.  93. 


Let  Pi  be  any  point  on  the  ellipse,  PoPs  any  diameter. 
Then  P^Pa  and  P1P3  are  supplemental  chords.  Let  the  coor- 
dinates of  Pi  be  (xi,  Yi)  and  those  of  P2,  (x.,,  y2).  Then  the 
coordinates  of  P3  are  ( —  X2,  —  ^2)^  since  P3  is  diametrically 
opj)osite  to  P2.  Then,  if  mi  and  mg  are  the  slopes  of  PiPo 
and  P1P3,  respectively,  we  have,  by  [2],  §  3, 

yi  — y2 

mi  — — 


m2 


mim 


Xi          X2 

_  yi  +  y2 

Xl   +  X2' 

^  _yi^-y2^ 

Hence 

Xi^ —  Xj" 

Now,  since  Pi  and  P2  are  botli  on  the  ellipse, 

Xi 


and 


—  + 
a^  ^  b- 


a^  "^  b^ 


yi^_ 


1, 


212  PLANE    ANALYTIC    GEOMETRY. 

By  subtraction, 

3?       ^        b^       ~"' 
or,  otherwise  written, 

yi^  —  ^2^  _       b^ 

Comparing  this  with  (1),  we  have,  finally, 

b^ 

mitTia^ ^, 

a 

which  proves  the  proposition. 

104.  Given  the  Extremity  of  any  Diameter,  to  find  the 
Extremities  of  the  Conjugate  Diameter. 

Let  Pi(xi,  Yi)  (Fig.  92)  be  the  extremity  of  a  diameter  P1P2 ; 
to  find  the  points  Qi  and  Q.,  at  the  extremities  of  the  conju- 
gate diameters. 

The  equation  of  CPi  is 

y  =  ~x,  (1) 

Xl 

and  therefore  the  equation  of  QiQo  is,  by  [39],  §  100, 

y  =  -^-  (2) 

a  \/i 


This  line  meets  the  ellipse 


/2 


-+y=i 

a^  ^  b^      -"' 
in  points  the  abscissas  of  which  are  given  by 

x_2       bV    2_. 

a^^aV     "~ 
Solving,  we  have 

„2_       ^  yi 


a^i'+b^xi^' 
But  since  (xj,  yi)  is  on  the  ellipse, 

b-xi^  +  a'^yr  =  a'^b^. 


(3) 


THE    ELLIPSE.  213 

Hence,  substituting  in  (3), 

„2     ^  yi 


whence  x  ^  ^r  -  yi. 

The  corresponding  values  of  y,  found  from  (2),  are 

b 
y  =  ±-x,. 

The  coordinates  of  Qi  and  Q^  are,  therefore, 

In  solving  numerical  examples  the  student  should  follow  the  methods 
of  this  article  and  not  use  the  results.     For  example,  given  the  ellipse 

^  +  y"  =  1 
9        4 

find  the  diameter  through  the  point  ('/,  |),  and  the  extremities  of  its 
conjugate  diameter. 

The  diameter  through  (y-,  |)  is 

1 

y  =  ti^^ 

'5' 

or  y  =  ix, 

with  slope  mi  =  h.    The  slope  m2  of  the  conjugate  diameter  is  given  by 

2  "2  a  > 

whence  m2  =  —  |. 

The  equation  of  the  conjugate  diameter  is,  therefore, 

y  =  -  'x- 
To  find  the  intersection  with  the  ellipse 

^4.y!-  1 
9        4~    ' 

we  solve  the  two  equations  with  the  result 

105.     Lengths  of  Conjugate  Diameters. 

We  will  denote  the  length  of  semi-diameter  CPi  (Fig.  92) 
by  a',  that  of  the  semi-diameter  CQi,  conjugate  to  CPi,  by  b'. 


214  PLANE   ANALYTIC    GEOMETRY. 

Let  the  coordinates  of  Pi  be  (xj,  yi) ;  then  those  of  Qi  are 


By[l],§3, 


Adding, 


a        b 
b        a 

a«=xr  +  y,»,  (1) 

b°  =  ^x.'  +  ^y.'-  (2) 

But,  since  Pi  is  on  the  ellipse, 

a-  ^  b--^ 
Hence 

a'--^  +  b"  =  a^  +  b^. 

We  may  state  this  result  in  the  following  theorem  : 
The  su7ii  of  the  squares  of  two  conjugate  semi^diameters  is 
constant  and  equal  to  the  sum  of  the  squares  of  the  semi-axes. 

106.     Angle  between  Conjugate  Diameters. 

Let  Pi  Pa  and  Q1Q2  (Fig-  92)  be  two  conjugate  diameters; 
required  to  find  the  angle  ^  between  them.  Draw  the  tan- 
gent PiN,  and  the  perpendicular  from  the  centre  upon  it.  As 
in  the  previous  article,  denote  the  length  of  CPi  by  a',  and 
that  of  CQi  by  b'.  Then,  by  §  102,  PiN  is  parallel  to  Q1Q2 
and,  hence,  the  angle  CPiN  equals  ^.     Therefore, 

CN        CN 
^^"'^^CP^^V 

It  remains  to  find  the  value  of  CN.  Now  the  equation  of 
the  tangent  PiN  is 


THE    ELLIPSE. 


215 


a-         b" 
or,  cleared  of  fractions, 

b'-XiX  +  a'Yiy  =  a'^b". 
By  §  32,  the  length  of  the  normal  from  the  centre  upon 
this  line  is 


which  may  be  written 


Vb^Xi^  +  aV 
ab 


zr  b'^ 

But  by  (2j  of  the  preceding  article,  the  quantity  under  the 
radical  sign  is  b'-. 

Therefore,  CN=^- 


Hence 


sin  ^ 


a'b' 


107.     Parallelogram  on  Conjugate  Diameters. 


Consider  the  parallelognun  GHKL  formed  by  drawing  tan- 
gents at  the  extremities  of  a  pair  of  conjugate   diameters, 


216 


PLANE    ANALYTIC    GEOMETRY. 


PiPo  and  QiQ,.  Since  (§102)  GL  and  HK  are  parallel  to 
Q1Q2  and  KL  and  GH  parallel  to  PiPo,  and  since,  furthei-, 
C  bisects  PiPo  and  Q1Q2,  it  follows  tliat  tlie  parallelograms 
PiCQoL,  PiCQiG,  P2HQ1C,  and  PgCQ.K  are  equal.  Now,  the 
area  of  CP1LQ2  is,  by  Trigonometry, 

CPi.CQo  sin  <^, 
=  a'b'  sin  <^, 

=  ab.  (§  106). 

Therefore,  Area  G K LH  =  4ab. 

Hence  : 

The  area  of  the  parallelogram  formed  by  dratvlng  tangents 
at  the  extremities  of  any  pair  of  conjugate  diameters  is  con- 
stant and  equal  to  the  rectangle  formed  by  drawing  tangents 
at  the  extremities  of  the  axes. 

108.     Auxiliary  Circle. 


Fig.  95. 


The  circle  described  on  the  major  axis  of  an   ellipse  as  a 
diameter  is  called  the  major  auxiliary  circle  of  the  ellipse.     If, 


THE    ELLIPSE.  217 

now,  any  ordinate  N  Pi  is  prolonged  until  it  cuts  the  auxiliary 
circle  in  a  point  R^,  then  P^  and  Ri  are  called  corresponding 
points.  It  is  evident  that  the  abscissas  of  corresponding  points 
are  the  same,  but  the  ordinates  differ.  There  is,  however,  a 
very  simple  relation  between  the  ordinates  NPi  and  NRj, 
which  may  be  found  as  follows.  Call  the  coordinates  of 
Pi  (xi,  yi)  and  those  of  Ri  (xj,  yg). 
Then,  since  Pi  is  on  the  ellipse 

a^  ^  b^         ' 
b 


we  have  Vi  ^  ^  Va''' 

a 

and  since  Ri  is  on  the  circle 

x'  +  y-  =  a^ 


we  have  ya  =  Va^'  —  Xi"  ; 

whence  follows  ^  =  -. 

y,       a 

That  is,  the  ordinate  of  any  2^oint  on  the  elVqjse  is  to  the  07'di- 
nate  of  the  conxspondiny  p)o^^^^  ^f  ^^^^  major  auxiliary  circle  as 
b:a. 


109.     Eccentric  Angle. 

I'he  eccentric  angle  of  a  point  on  an  ellipse  is  the  angle  made 
tvith  the  major  axis  by  the  line  joining  the  centre  to  the  corre- 
sponding point  of  the  auxiliary  circle. 

In  Fig.  95,  then,  NCRi  is  the  eccentric  angle  of  the  point  Pi. 
The  coordinates  of  a  point  are  readily  expressed  in  terms 
of  its  eccentric  angle.     For,  in  tlie  right  triangle  NCRi,  de- 
noting the  Z  NCRi  l^>y  </>'  we  have 

CN^CRiCos  <^. 
But  CN  =  xi  and  CRi  =  a. 

.*.  Xj  =  a  cos  <^. 


218 
Also, 

But,  by  §  108, 


PLANE    ANALYTIC    GEOMETRY. 


NRi=CRi  sin  <^ 
=  a  sin  <fi. 


.'.  Yi  =  b  sin  <^. 
The  required  relations  are,  therefore,  dropping  the  suffixes. 


X  =  a  cos  (|), 
y  =  b  sin  <j). 


[41] 


y^_ 


The  two  equations   [41]   are  together   equivalent   to   the 
single  equation 

a'  '    b^ 

This  may  be  seen  by  dividing  the  first  by  a,  the  second  by  b, 
and  then  squaring  and  adding. 


n.-fi.-l- 


110.     Eccentric  Angles  and  Conjugate  Diameters. 


— X 


Fig.  9G. 


THE    ELLIPSE.  219 

Let  CPi  and  CQi  be  two  conjugate  semi-diameters, 
/_  NCRi  =  ^,  the  eccentric  angle  of  P^  and  A  NCSi  =  <^', 
the  eccentric  angle  of  Qi.  Then,  if  the  coordinates  of  P^  are 
(xi,  yi)  and  those  of  Qj  are  (xo,  y2),  the  equation  of  CPi  is 

yi 

Xi 


ya 

y  =  —  X. 

Xo 


mi  = 

^yi 

b 

sm 

<^ 

a 

cos 

«A' 

3i, 

m2  = 

X2 

= 

b 
a 

sm 
cos 

b^ 

sin 

^ 

sin  <^' 

(by  [41]) 


and  that  of  CQi  is 

Hence,  for  the  slope  of  CP 
mi 
and,  similarly,  for  CQi, 

Therefore, 

a"  cos  (f>  cos  cf>' 
But  mim^  =  -  K.,  by  [40],  §  101. 

3," 

Therefore,  sin  <^  sin  <^  _  _ 

cos  (ft  cos  cji' 

or  cos  c}>  cos  <f>'  +  sin  <f>  sin  <f>'  =  0. 

This  may  be  written,  by  Trigonometry, 

cos  (<^'  —  <^)  =  0, 

and,  therefore,  ^'  —  <^  =  -. 

Prom  this  follows  the  theorem  : 

The  eccentric  angles  of  the  extremities  of  a  pair  of  conjugate 
diameters  differ  hy  90°. 

This  proposition  may  be  used  to  prove  the  propositions  of 
§§  104  and  105.  For  example  (§  104),  if  Xi,  yi  are  given,  we 
find  by  [41] 


220  PLANE   ANALYTIC   GEOMETRY. 

Xi 

COS  <b  =  -—, 

a 

sin  <}>  =  P-- 
Then  for  Qi, 
X2  =  a  cos  cji'  ^  a  cos  (^  +  75)  =  —  a  sin  ^  =  —  -  yi 

y2  =  b  sin  <^'  ^  b  sin  [  ^  +  ^  ]  =  b  cos  <^  =  -  Xi. 


Similarly,  for  Q^, 

X3  =  a  cos  ^"  =  a  cos  (  (j>  —  -\  =^  a  sin  (f>  =  -  yi, 

yg  =  b  sin  </>"  :=  b  sin  (  (f>  —  -j=  —  b  cos  <j)^= Xj. 

Also,  to  prove  §  105,  by  this  method,  we  have 

a'-  =  xi^  +  yi^  =  a^  cos^  ^  +  b^  sin^  <f>, 
b'-  =  xo^  +  y2^  =  a^  sin^  <^  +  b"  cos^  <^. 

By  adding,  remembering  that  sin-  (f>  -\-  cos'-'  ^  =:  1,  we  get 
a'2  +  b'2  ==  a-  +  b-. 

111.     Ellipse  referred  to  Conjugate  Diameters  as  Axes. 

Let  it  be  required  to  transform  the  equation  of  the  ellipse 

a-       b^ 

from  the   old  axes  OX  and  OY  to  new  ones   OX'  and  OY', 
the  latter  being  a  pair  of  conjugate  diameters  of  the  ellipse. 

Since  the  transformation  is  from  rectangular  axes  to 
oblique  axes  having  the  same  origin,  we  have,  by  [21],  §  45, 
the  formulas  of  transformation, 

x  =  x'  cos  ^  +  y'  cos  6', 
y  ^  x'  sin  6  -\-  y'  sin  6'. 


THE    ELLIPSE. 


■221 


FlfJ.  97. 


Substituting  in  the  equation  of  the  ellipse,  we  have 

(x'  cos  ^  +  y'cos  O'y      (x'  sin  6)  +  y'  sin  6')-  _ 

a^  +  b^  ~^' 


which  reduces  to 

cos^  6   ,   sin^ 


,    „  ,  ,  ,  cos  0  cos  6'  ,   sin  6  sin  6' 
+  2x'y    •;  -, h 


=  1. 


,     ,„  ,  ccs'^  0'   ,   sin^  e' 


We  need  now  to  see  what  special  form  the  coefhcients 
assume  owing  to  the  hypothesis  that  CX'  and  CY'  are  con- 
jugate diameters. 

In  the  first  place,  the  relation 

b^ 

is  the  same  as 

b^ 


tan  0  tan  0'  =  — 


which  is  equivalent  to 

cos  6  cos 


+ 


sm  a  sm 


a'  b^ 

Hence  the  coefficient  of  x'y'  vanishes. 


0. 


222  PLANE    ANALYTIC    GEOMETRY. 

Moreover,  if  we  call  CPi  =  a',  as  before,  and  take  the  coor- 
dinates of  Pi,  referred  to  the  old  axes,  as  Xi,  y^,  we  have 

cos  6^-„  sin  6  =  ^. 
a  a' 

The  coefficient  of  x'^  is,  therefore, 

1/xi^       yA 

a'2  i^a^"^  b-/ 

which  equals  simply  — ,  since  Pi  is  on  the  ellipse.     Similarly, 
a 

the  coefficient  of  y'-  may  be  proved  to  be  — .     The  equation 

of   the  ellipse  referred  to  the  new  axes  is,  therefore,   the 
primes  on  the  coordinates  being  dropped. 

It  will  be  noticed  that  this  is  an  equation  of  the  same  form 
as  the  ori^rinal  one. 


112.     Polar  Equation  of  the  Ellipse. 

To  find  the  polar  equation  of  the  ellipse,  the  pole  being  at  the 
centre. 

To  transform  from  rectangular  coordinates  to  polar  coordi- 
nates, we  have  the  relation  [7],  §  10, 
X  ^  r  cos  Q, 
y  =  r  sin  Q. 

These  values,  substituted  in 


a^  ^  b^      -^' 

l''  / 

^cos=^^  ,  sin2^\ 

'  I 

,  a'  +  b'  y 

a=b' 

give  r^  f  — ^^  -f  -^j-  ]  ==  1, 

whence  .  — ,  „       „  .  ,     ,    •  2  ^' 

b^  cos^  Q-\-  z?  sm^  Q 


THE  p:llii\se.  223 


a^— (a^— b^)  cos^^' 


This  is  the  required,  equation. 


■•■'•'=  l-e-cs'e'  W3J 


EXAMPLES. 

1.  Given  the  ellipse  3x-  +  5y-  =  1  ;  find  the  semi-axes, 
eccentricity,  foci,  directrices,  and.  latus  rectum. 

2.  Find  the  equation  of  an  ellipse  which  passes  through 
the  points  (1,  —  2)  and  (3,  —  1),  the  centre  of  the  ellipse 
being  at  the  origin,  and.  the  axes  of  tlie  ellipse  coinciding 
with  the  axes  of  coordinates. 

3.  Find  the  equation  of  an  ellipse,  the  eccentricity  of  which 
is  f  and  the  foci  of  which  lie  at  the  points  (0,  ±  6). 

4.  Find  the  equation  of  an  ellipse,  the  eccentricity  of  which 
is  §  and  the  latus  rectum  10,  the  centre  of  the  ellipse  being 
at  the  origin,  and  the  axes  of  tlie  ellipse  coinciding  with  the 
axes  of  coordinates. 

5.  Does  the  point  (1,  3)  lie  inside,  on,  or  outside,  the 
ellipse  10x'+lly2=110? 

6.  Find  the  eccentricity  and  the  equation  of  an  ellipse,  if 
the  foci  lie  halfway  between  the  centre  and  the  vertices. 

7.  Find  the  eccentricity  and  the  equation  of  the  ellipse,  if 
the  latus  rectum  is  ^  the  minor  axis. 

8.  Find  the  eccentricity,  if  the  line  connecting  the  positive 
extremities  of  the  axes  is  parallel  to  the  line  joining  the 
centre  to  the  upper  end  of  the  left-hand  latus  rectum. 

9.  Find  the  equation  of  the  tangent  and  normal  to  the 
ellipse  36x-+  108y"^20  at  a  point  the  abscissa  of  wliieh  is  ^. 

10.  Find  the  lengths  of  the  subtangent  and  the  subnormal 
for  the  point  (2,  —  3)  on  the  ellipse  4x-  -j-  7y^  =  79.  (See 
definition,  §  56.) 


224  PLANE    ANALYTIC    GEOMETRY. 

11.  Find  the  tangent  and  the  normal  to  the  ellipse 

at  the  upper  extremity  of  the  right-hand  latus  rectum. 

12.  Find  the  tangents  to  the  ellipse  4x^  +  9y^  =  36  which 
are  parallel  to  the  line  joining  the  positive  extremities  of  the 
axes. 

13.  Find  the  tangents  to  the  ellipse  2x^  +  5y^  =  10  which 
make  an  angle  of  60°  with  the  axis  of  x. 

14.  Find  the  point  in  which  the  polar  of  (4,  5)  with  respect 

to  the  ellipse  —  +  =^  =  1  intersects  the  straight  line  joining 

the  pole  to  the  origin. 

15.  Find  the  equations  of  the  tangent  and  normal  at  the 

upper  extremity  of  the  left-hand  latus  rectum  of  the  ellipse 

x^       y2 

— +  r^=lj  in  terms  of  the  major  axis  and  the  eccentricity. 

16.  If  the  normal  in  Ex.  15  passes  through  the  upper  end 
of  the  minor  axis,  show  that  the  eccentricity  is  given  by  the 
equation  e*  +  e^  —  1=0. 

o  9 

x"       V 

17.  Find  a  point  on  the  ellipse  —  +  f^  =  1?  such  that  the 

tangent  there  is  equally  inclined  to  the  two  axes. 

18.  Find  expressions  for  the  subtangent  and  the  subnormal 

for  a  point  (xi,  yi)  on  the  ellipse  "^  +  ri  =  1- 

cl  D 

19.  Show  that  the  point  (xj,  yi)  lies  inside,  on,  or  outside, 

x^      y2 
the  ellipse  — 2  +  ui  ^^  1'  according  as 

|-  +  ^^-l<,  =  ,o,>0. 

20.  Prove  analytically  that  if  the  normals  at  all  points  of 
an  ellipse  pass  through  the  centre,  the  ellipse  is  a  circle. 

21.  Find  the  coordinates  of  a  point  such  that  the  tangent 
there  is  parallel  to  the  line  joining  the  positive  extremities  of 
the  major  and  the  minor  axes. 


THE    ELLIPSE.  225 

22.  Find  the  equation  and  the  length  of  the  perpendicular 
from  the  centre  to  any  tangent  to  the  ellipse 

23.  Show  that  the  product  of  the  perpendiculars  from  the 
foci  upon  any  tangent  equals  the  square  of  half  the  minor 
axis. 

24.  Show  that  if  any  ordinate  M  P  be  produced  to  meet  the 
tangent  at  the  end  of  the  latus  rectum  through  F  in  Q,  then 
QM  =  FP. 

25.  Show  that  the  perpendicular  from  the  focus  upon  any 
tangent  meets  the  line  drawn  from  the  centre  to  the  point  of 
contact  on  the  corresponding  directrix. 

26.  Prove  that  any  choxd  through  the  focus  is  perpen- 
dicular to  the  line  joining  its  pole  to  the  focus. 

27.  Find  the  locus  of  the  middle  points  of  a  system  of 

y2  y2 

chords  of  the  ellipse   —  +  ^  =  1,  which  are  parallel  to  the 

line  2x  -f-  3y  —  1  =  0. 

28.  In  the  ellipse  2x-  +  5y^  =  3,  find  two  conjugate  diam- 
eters, one  of  which  bisects  the  chord  x  +  3y  +  2  ^  0. 

29.  In  the  ellipse  2x^  +  3y^  =  l,  find  two  conjugate  diam- 
eters, one  of  which  passes  through  (^,  |). 

30.  In  the  ellipse  4x2  _^  25y^  =  100,  find  two  conjugate 
diameters,  one  of  which  is  perpendicular  to  3x  —  4y  +  8  =  0. 

31.  Given  the  point  (—  2,  3)  on  the  ellipse  x-  +  2y=^  =  22, 
find  the  conjugate  points. 

32.  In  the  ellipse  3x2  +  4y"^  =  12,  g,-^(j  ^]^g  equation  of  a 
chord,  the  middle  point  of  which  is  (2,  —  1). 

33.  In  the  ellipse  -,-\-h  —  '^,  finfl  the  equations  of   two 

S"         D 

conjugate  diameters,  one  of  which  bisects  the  chord  deter- 
mined by  the  upper  end  of  the  minor  axis  and  the  right-hand 
focus. 


226  PLANE   ANALYTIC    GEOMETRY. 

34.  In  any  ellipse,  show  that  the  diameters  parallel  to  the 
lines  joining  the  extremities  of  the  axes  are  conjugate. 

35.  Show  that  the  tangents  at  the  extremities  of  any  chord 
meet  on  the  diameter  which  bisects  the  chord. 

36.  If  the  tangent  at  the  vertex  A  cut  any  two  conjugate 
diameters  produced  in  T  and  t,  show  that  AT.  At  =  —  b^ 

37.  Show  that  the  product  of  the  focal  distances  of  any 
point  is  equal  to  the  square  of  half  the  corresponding  conju- 
gate diameter. 

38.  If  from  the  focus  of  an  ellipse  a  perpendicular  is  drawn 
to  a  diameter,  show  that  it  will  meet  the  conjugate  diameter 
on  the  corresponding  directrix. 

39.  Find  the  diameter  of  the  ellipse 

16^  4 

which  is  equal  to  its  conjugate  diameter. 

40.  Show  that  there  can  be  only  one  pair  of  equal  conju- 
gate diameters  of  the  ellipse 

x^      y^  b  b 

^  +  ^2  ""  ^  '  namely,  y  =  -x,  y  =  --x. 

41.  Given  the  point  (—  3,  —  1)  on  the  ellipse  x^  +  3y2  =  12  ; 
find  the  corresponding  point  on  the  auxiliary  circle,  and  the 
eccentric  angle. 

x^      y2 

42.  In  the  ellipse  -  +  j  =  1,  find  the  point  of  which  the 

eccentric  angle  is  30°. 

43.  Show  that  the  tangent  to  the  ellipse  at  (xj,  yj)  and  the 
tangent  at  the  corresponding  point  of  the  auxiliary  circle  pass 
through  the  same  point  of  the  major  axis. 

44.  Find  the  equation  of  the  tangent  at  any  point  of  an 
ellipse  in  terms  of  the  eccentric  angle  at  that  point. 

45.  Show  that  the  area  of  a  triangle  inscribed  in  an  ellipse 
is  ^  ab  [sin  (/8  —  y)  +  sin  (y  —  a)  +  sin  (a  —  ^)],  where  a,  (3,  y 
are  the  eccentric  angles  of  the  vertices  of  the  triangle. 


THE    ELLIPSE.  227 

46.  Show  that  the  perpendicular  from  either  focus  upon 
the  tangent  at  any  point  of  the  auxiliary  circle  equals  the 
focal  distance  of  the  corresponding  point  of  the  ellipse. 

47.  Show  that  the  eccentric  angles  of  the  extremities  of 

two  equal  conjugate  diameters  are  -r  and  —-• 

48.  Find  the  equation  of  the  ellipse  2x^  +  4y^=  12  referred 
to  a  pair  of  conjugate  diameters  as  axes,  one  of  which  passes 
throiigh  the  point  (2,  1). 

49.  Find  the  equation  of  the  ellipse  2x^  +  4y'  =  12  referred 
to  its  equal  conjugate  diameters  as  axes. 

50.  Show  that  the  equation  of  any  ellipse  referred  to  its 

equal  conjugate  diameters  as  axes  is  x^  +  y^= — - — . 

51.  If  A  and  A'  be  the  extremities  of  the  major  axis  of  an 
ellipse,  T  the  point  where  the  tangent  at  any  point  P  meets 
AA',  QR  a  line  through  T  perpendicular  to  AA'  and  meeting 
AP  in  Q  and  A'P  in  R,  show  that  QT  =  RT. 

52.  From  a  point  P  on  an  ellipse  straight  lines  are  drawn 
to  A  and  A',  and  from  A  and  A'  straight  lines  are  drawn  per- 
pendicular to  AP  and  A'P.  Show  that  the  locus  of  their  point 
of  intersection  is  an  ellipse. 

53.  Find  the  locus  of  the  points  of  intersection  of  normals 
at  corresponding  points  of  the  ellipse  and  the  auxiliary  circle. 

54.  Find  the  locus  of  the  vertices  of  a  parallelogram 
formed  on  a  pair  of  conjugate  diameters. 

55.  Chords  are  drawn  througli  the  end  of  the  major  axis  of 
the  ellipse  ;  find  the  locus  of  their  middle  points. 

56.  Find  the  locus  of  the  middle  points  of  a  system  of 
chords  passing  through  a  fixed  point  (xi,  y^). 

57.  An  ellipse  moves  so  as  to  be  always  tangent  to  a  pair 
of  rectangular  axes;  show  that  the  locus  of  its  centre  is  a 
circle. 

58.  A  point  moves  so  that  the  sum  of  the  squares  of  its 
distances    from  two    intersecting   straight  lines  is  constant. 


228  PLANE    ANALYTIC    GEOMETRY. 

Prove  that  the  locus  is  an  ellipse,  and  find  its  eccentricity  in 
terms  of  the  angle  between  the  lines. 

59.  Two  tangents  to  an  ellipse  are  so  drawn  tliat  the  prod- 
uct of  their  slopes  is  constant.  Show  that  the  locus  of  their 
point  of  intersection  is  an  ellipse  or  hyperbola  according  as 
the  product  is  negative  or  positive. 


CHAPTER    X. 


^  =  1. 


113.     Conjugate  and  Equilateral  Hyperbolas. 

We  have  seen  in  §  69  that  the  equation  of  the  hyperbola 
takes  its  simplest  form 

S-^.=i.  (1) 

when  the  centre  of  the  curve  is  taken  as  the  origin  of 
coordinates   and   the    transverse    axis    of   the    curve    as    the 

VaM-^ 
axis   of    X.      The  eccentricity  is   e  = ;    the  foci  lie 

£1 

at  the  points  (±  ae,  0)  ;  and  the  equations  of  the  directrices 

are  x  ^  rb  - .     The  polar  of  the  point  (x,,  y,)  is  -^  —  ^~  =  1, 
6'  ^        a''        b- 

this  equation  representing  the  tangent  as  special  case  of  the 
polar  when  (xi,  yi)  is  on  the  curve.  If  the  slope  m  of  a  tan- 
gent is  given,  its  equation  is  of  the  form,  y  =  mx  ±  Va"^in- —  b". 
Closely  connected  Avitli  this  hyperbola  is  the  one  the  equa- 
tion of  which  is 

where  a  and  b  have  the  same  numerical  values  as  in  the  first 
equation.  The  hyperbolas  (1)  and  (2)  are  so  related  that 
the  transverse  axis  of  each  is  the  conjugate  axis  of  the 
other;  hence  they  are  included  between  the  same  asj^mptotes 

y  =  ±  -  x.     They  lie,  therefore,  as  in  Fig.  98,  the  hyperbola 


230 


PLANE    ANALYTIC    GEOMETRY. 


(1)  being  composed  of  the  right  and  left-hand  branches,  and 
the  hyperbola  (2)  of  the  upper  and  lower  branches. 


Such  hyperbolas  are  said  to  be  conjugate  to  each  other,  and 
in  particular  (1)  is  usually  called  the  primary  hyperbola  and 
(2)  the  conjugate  hyperbola  * 

The  eccentricity  of  the  conjugate  hyperbola  is 

its  foci  lie  at  the  points  (0,  ±  be');  and  the  equations  of  its 

,.  .  b 

directrices  are  y  =  ±  — . 
e 

Evidently   two    conjugate    hyperbolas    are,   in   general,  of 

quite  different  shapes,  and  could  not  be  made  to  coincide. 

An  exception   occurs   only  when  b  ^  a.     In  tliat  case,  the 

equation  of  the  primary  hyperbola  is 


*  It  would  be  just  as  correct  to  call  (2)  the  primary  hyperbola,  and 
(1)  the  conjugate  hyperbola.  That  hyperbola  is  the  primary  one  which 
appears  first  in  the  discussion  of  a  problem. 


THE    HYPERBOLA.  231 

The  asymptotes  are  the  lines  y  =  ±  x,  which  bisect  the 
angles  between  the  axes  of  coordinates,  and  are  perpendicular 
to  each  other.  Sucli  an  hyperbola  is  called  an  equilateral  or 
rectangular  hyperbola;  the  lirst  name  being  derived  from 
the  equality  of  the  axes;  the  second,  from  the  angle  between 
the  asym})totes.     The  conjugate  hyperbola  is  then 

—  x^  +  y^  =  a.', 

which  is  equal  to  the  primary  hyperbola,  but  turned  through 
an  angle  of  90°. 

In  this  chapter,  we  shall  work  with  the  primary  hyperbola 

a-       b- 

In  those  cases  where  the  conjugate  hyperbola  is  used,  ex- 
plicit mention  will  be  made  of  the  fact.  The  special  form  of 
the  result  for  the  equilateral  hyperbola  may  be  obtained  by 
placing  b  =:  a,  but  we  will  omit  this  in  all  cases. 

The  student  will  notice  that  much  of  the  following  work  is 
similar  to  that  of  tlie  preceding  chapter.  In  nuiny  cases  tlie 
results  differ  only  in  a  sign,  corresponding  to  the  fact  that 
the  equations  of  the  ellipse  and  the  hyperbola  differ  only  in 
that  way. 

114.     To  find  the  Foci  when  the  Axes  are  given. 

From  §  69,  the  distance  of  the  foci  from  tlie  centre  is  ae, 
which  equals  Va^  +  b^.  This  equals  the  lengtli  of  the  semi- 
diagonal  CH  of  the  rectangle  on  the  axes  (Fig.  98).  The 
distance  of  the  foci  of  the  conjugate  hyperbola  from  the 
centre  is  be',  which  also  equals  Va-  -|-  b". 

Hence  the  following  rule  : 

From  the  centre  of  the  hijperhohi  describe  a  circle  with  a 
radius  equal  to  the  semi-diagonal  of  the  rectangle  on  the  axes. 
The  circle  will  cut  tlie  transverse  axis  in  the  foci  of  the  prlnnrri/ 


232 


PLANE    ANALYTIC    GEOMETRY. 


hyperbola,  and  the  conjugate  axis  in  the  foci  of  the  conjugate 
hyperbola. 


115.     Focal  Distances. 

The  definitions  of  focal  distances  and  focal  radii  are  the 
same  as  in  the  case  of  the  ellipse.  To  investigate  we 
proceed  as  in  §  94. 

Y 


Let  C  be  the  centre  of  the  hyperbola,  F  and  F'  the  foci,  and 
DM  and  D'M'  the  directrices.  Take  Pi  a  point  on  the  right- 
hand  branch  and  draw  the  ordinate  NPi,  the  focal  radii  F'Pi 
and  FPi,  and  the  line  P^MM'  perpendicular  to  the  directrices. 
Then  by  defijiition  of  the  hyperbola 
FPi=re.  MPi, 

,  =  e.(CN-  CD), 


=  e     x,  — 


Also, 


^  ex,  —  a. 
F'Pi  =  e.  M'P„ 

=  6.  (CN  +  D'C), 

=  e  (  xi  -f 


=  exi  -)-  a. 


THE    HYPERBOLA.  233 

Similarly,  for  a  point  on  the  left-hand  branch,  we  find 
F'Pi  =  —  exi  —  a, 
and  FPi  =  —  exi  +  a. 

116,  Difference  of  Focal  Distances. 

From  the  results  of  the  last  article,  we  find 
F'Pi-FPi  =  2a 
for  the  right-hand  branch,  and 

FPi-  F'Pi  =  2a 
for  the  left-hand  branch.     Hence  : 

The  difference  between  the  focal  distances  of  any  point  on  an 
hypevhola  is  constant  and  equal  to  the  transverse  axis. 

This  property  is  true  for  no  point  not  on  the  hyperbola,  as 
may  be  readily  proved  by  use  of  the  axiom  that  a  straight 
line  is  the  shortest  distance  between  two  points.  Hence  we 
may  say  : 

The  liyj^erhola  is  the  locus  of  a  j^oint,  the  difference  of  the 
distances  of  which  from  two  fixed  points  is  constant. 

Tliis  is  sometimes  given  as  the  definition  of  the  hyperbola 
and  is  made  the  basis  of  the  investigation  of  its  equation  and 
properties. 

117.  Constructions. 

The  preceding  article  forms  the  basis  of  the  following 
methods  of  constructing  an  hyperbola  when  the  two  axes,  or 
the  foci  and  the  transverse  axis,  are  given. 

(1)  By  means  of  compasses. 

If  the  foci  are  not  given,  find  them  as  in  §  114.  Then, 
take  a  straight  line  equal  to  the  transverse  axis,  and  divide  it 
externally  in  any  two  portions  r  and  r'. 

With  one  focus  as  a  centre  and  r  as  a  radius,  describe  an 
arc  of  a  circle  ;  with  the  other  focus  as  a  centre  and  r'  as 


234 


PLANE   ANALYTIC    GEOMETRY. 


radius,  describe  another  arc.  The  intersections  of  the  two 
arcs  are  points  on  the  hyperbola.  In  this  way  any  number 
of  points  may  be  found  and  the  curve  drawn. 

(2)  By  a  mechanical  device. 

First  find  the  foci,  if  they  are  not  given.  Then  take  a 
ruler  longer  than  the  transverse  axis,  and  a  shorter  thread 
sucli  that  the  difference  in  length  of  the  ruler  and  the  thread 
shall  equal  the  transverse  axis.  Pivot  one  end  of  the  ruler 
at  one  of  the  foci,  and  attach  one  end  of  the  thread  to  tlie 
free  end  of  the  ruler  and  the  other  end  of  the  thread  to  the 
unused  focus.  Insert  a  pencil  in  the  loop  of  the  thread  and 
press  it  tightly  against  the  ruler.  As  the  ruler  is  revolved 
around  its  pivot,  the  pencil  describes  one  branch  of  the 
hyperbola.  By  interchanging  the  positions  of  the  ruler  and 
the  thread,  the  other  branch  of  the  hyperbola  is  obtained. 

118.     Angle  between  Tangent  and  Focal  Radii. 

The  tangent  bisects  the  angle  hetiveen  the  focul  radii  of  the 
jioint  of  contact,  and  the  normal  makes  eqital  angles  with  these 
focal  radii. 


Fig.  100. 


THE    HYPERBOLA.  235 

Let   PiT  be  the  tangent  at  any  point   Pi(xi,  yi),  P^R  the 
normal,  and  FPi  and  F'Pi,  the  focal  radii. 
We  are  to  prove 

ZF'PiT=ZTPiF, 

and  ZFPiR  =  ZRPiQ. 

The  equation  of  the  tangent  is 

xix  _  yiy  ^ 
a^        b^         ' 

whence  follows  that  CT,  the  intercept  on  the  axis  of  x, 

a^ 


Therefore 


FT 

^1 
=  F'C  +  CT, 

1   ^' 
=  ae  H > 

Xi 

aexi  +  a^ 

Xi 

TF 

-:CF-CT, 

a^ 
=  ae > 

Xl 

aexj  —  a" 

(§  115) 


and 


Xl 

FT  _  aexi  +  a- _  exi  +  a F'Pi 

TF       aexi  —  a^      exj  —  a       FPj 

Hence,  by  Geometry, 

ZF'PiT  =  ZTPiF. 

Also,  PiR  is  by  definition  perpendicular  to  PjT. 

.-.  ZFPiR  =  ZRP,Q. 

The  value  we  have  found  for  CT  sliows  that  a  tangent  to  the  right- 
hand  branch  of  the  hyperbola  cuts  the  transverse  axis  at  the  right  of  the 
centre,  and  a  tangent  to  the  left-hand  branch  cuts  the  transverse  axis  at 
the  left  of  the  centre.  A  reference  to  this  fact  will  .sonietimes  save  the 
student  from  false  constructions. 


236 


PLANE    ANALYTIC    GEOMETRY. 


From  §  18,  any  straight  line  cuts  a  conic  in  no  more  than  two  points, 
and  a  point  of  tangency  is  counted  as  two  coincident  points.  Hence  the 
tangent  to  one  branch  does  not  cut  the  other  branch. 


119.     Equation  of  a  Diameter. 

As  already  defined,  a  diameter  of  a  conic  is  the  locus  of 
the  middle  points  of  a  system  of  parallel  chords.  It  is  to  be 
noticed  that  parallel  chords  in  an  hyperbola  may  be  drawn  in 
two  ways.  They  may  connect  points  on  the  same  branch  or 
they  may  connect  points  on  different  branches,  as  shown  by 
the  two  systems  of  parallel  chords  in  Fig.  101. 


In  either  case,  by  using  the  method  of  §  61,  we  find  that 
the  equation  of  the  diameter  bisecting  the  system,  the  slope 
of  which  is  rrii,  is 


y  = 


a-m 


X. 


[44] 


If  m2  is  tlie  slope  of  the  diameter,  then  the  relation  between 
m2  and  rrii  is 

mim2  =  -:i. 


THE    HYPERBOLA.  237 

From  this  follows  that  the  axes  are  the  only  diameters  per- 
pendicular to  the  chords  they  bisect.     For  m^  cannot  equal 

—  — ,  unless  mi  =  0  and  m.,  =  go,  or  vice  versdi; 
nrii 

120.     Conjugate  Diameters. 

As  in  the  ellipse,  conjugate  diameters  are  such  that  each 
bisects  all  chords  parallel  to  the  other.  It  readily  follows 
that  the  relation  between  their  slopes  is 

rriim.,  =  -,•  [45] 

a 

From  this  follow  the  theorems  : 

1.  The  two  axes  are  the  only  i^air  of  conjugate  diameters  that 
are  j^erpendicular  to  each  other. 

2.  Two  conjugate  diameters  make  either  both  actite  or  both 
obtuse  angles  with  the  transverse  axis. 

For  the  product  of  trii  and  nria  is  always  positive,  and  hence 
the  two  factors  have  the  same  sign.  Therefore  the  upper 
portions  of  two  conjugate  diameters  lie  upon  the  same  side 
of  the  conjugate  axis  ;  for  example,  see  P1P2  and  QiQ-  in 
Fig.  101. 

3.  One  diameter  intersects  the  hyperbola,  the  other  does  not. 
For  a  line  y  ^  iriix  intersects  the  hyjjerbola  in  points  the 

abscissas  of  which  are 

ab 


Vb^  —  a'-rrii^ 
These  values  of  x  are  real  when  irii  is  numerically  less  than 

-,  and   imaginary  when   mi  is  numerically  greater  than   -. 
a  a 

But  of  two  conjugate  diameters  the  slope  of  one  is  always 

numerically  greater  than  -  and  the  other  less,  by  virtue  of  the 
a 

relation    mim.,  =  — . 
a' 


238  PLANE   ANALYTIC    GEOMETRY. 

An  exception  occurs  only  when  rrii  =  ±  -.     Then  rria  =  ±  - 

also,  and  the  two  diameters  coincide  with  each  other  and 
with  an  asymptote, 

4.  A  diameter-  intersects  either  the  primary  Jtyperhola  or  the 
conjugate  hyperbola. 

For  as  we  have  just  seen,  the  diameter, 

y  =  mjx, 
intersects  the  primary  hyperbola 

a^      b^ 

when  irii  is  numerically  less    than    -,    but   not   when  mi  is 

a 

numerically  greater  than  -.     In  the  latter  case,  however,  it 

a 

Avill  intersect  the  conjugate  hyperbola 

a^  ^  b^        ' 

for  the  abscissas  of  the  intersections  of  y  =  miX  and  this 
curve  are  given  by 

ab 


V—  b^  +  a-^m^ 
which  is  real  when  nrii  is  numerically  greater  than  -. 


a 


121.     Propositions  on  Conjugate  Diameters. 

The  following  propositions  are  similar  to  corresponding 
ones  for  the  ellipse.  We  state  them  and  leave  the  proof  to 
the  student. 

One  point,  however,  needs  a  little  explanation.  We  have 
seen  that  of  two  conjugate  diameters,  one  intersects  the 
hyperbola,  the  other  does  not.  The  one  appears,  therefore, 
indefinite  in  extent,  while  the  other  may  be  limited  by  the 


THE    HYPERBOLA.  239 

curve.  We  may  overcome  this  want  of  uniformity  by  remem- 
bering that  a  diameter  wliich  does  not  intersect  the  primary 
hyperbola  does  intersect  tlie  conjugate  liyperbola.  Henca  tve 
will  agree  to  call  the  extreinities  of  any  diameter  the  points  in 
which  it  cuts  either  the  i)riniary  or  the  conjugate  hijperhola ^  as 
the  case  may  be. 

Moreover,  we  may  remark  tliat  diameters  conjugate  with 
reference  to  the  primary  hyperbola  are  conjugate  with  refer- 
ence to  the  conjiigate  hyperbola.  For  we  may  change  the 
equation  of  the  primary  hyperbola  into  that  of  the  conjugate 
hyperbola  by  writing  —  a"  for  a^  and  —  b^  for  b^.  Such  a 
change,  however,  leaves  the  relation  [45]  unaltered. 

1.  The  tangent  at  tlie  extremity  of  any  diameter'  is  parallel  to 
the  conjugate  diameter.  (§  102) 

2.  Any  pair  of  supplemental  chords  is  2)arallel  to  a  pair  of 
conjugate  diameters.  (§  103) 

3.  If  (>ii,  yi)  is  the  extremity  of  any  diameter,  then 
a  b 
b^^'-r^ 

are  the  extremities  of  the  conjugate  diameter.  (§  104) 

4.  If  a'  and  h'  are  the  lengths  of  a  pair  of  conjugate  semi- 
diameters,  then 

a'2  -  b"-^  =■  a-  -  b\  (§  105) 

5.  If  (f)  is  the  angle  betiveen  a  2)uir  of  conjugate  diameters, 

sin<^  =  ^,-  (§106) 

6.  The  area  of  the  parallelogram  formed  by  tangents  at  the 
extremities  of  a  pair  of  conjugate  diameters  is  constant  and 
equal  to  the  rectangle  on  the  axes.  (§  107) 

7.  The  equation  of  the  hyperbola  referred  to  a  pair  of  conju- 
gate diameters  as  axes  of  coordinates  is 

^-^=1-  [46] 


240 


PLANE    ANALYTIC    GEOMETRY. 


122.     Asymptotes. 

We  have  already  noticed  in  §  69  that  the  two  straight  lines 

y 


b 
a   ' 


which  are  readily  constructed  as  the  diagonals  of  the  rect- 
angle on  the  axes,  possess  the  peculiar  property  of  having 
the  coordinates  of  their  points  of  intersection  with  the  hyper- 
bola expressed  as  infinite  quantities.  These  lines  Ave  call 
the  asymptotes  of  the  hyperbola. 

In  general  we  may  define  an  asymptote  of  a  curve  as  a 
straight  line  of  ivhich  the  distance  from  a  point  on  the  curve 
decreases  below  any  assignable  quantity  as  the  jmint  of  the 
curve  recedes  indefinitely  from  the  origin. 

We  proceed  to  show  that  the  asymptotes  of  the  hyperbola 
satisfy  this  definition.  For  that  purpose,  let  us  take  a  point 
Pi(xi)  Yi)  on  the  upper,  right-hand  portion  of  the  hyperbola, 
and  denote  its  perpendicular  distance  from  the  neighboring 
asymptote  by  PiR. 


Fig.  102. 

If  we  write  the  equation  of  the  asymptote  in  the  form 
bx  —  ay  =  0, 

_  bxi  — ayi 


then,  by  §  32, 


PiR 


Va^  +  b^ 


THE    HYPERBOLA. 


241 


,  since  (xi 

yi)  i« 

on  the  hyperbola, 

yi 

h 

-3  Vx;^-a^ 

ice,  by  sill 

)stituti 
=  b(x 

on. 

PiR 

—  Vxi'^  —  a-) 
Va'^  +  b- 



ba^ 

Va^  +  b^  (xi  +  V  xi-  -  a^) 

Now  as  Xi  is  allowed  to  increase  indefinitely,  the  value  of 
PjR  decreases  indefinitely,  as  was  to  be  proved. 

We  give  in  the  following  articles  a  few  of  the  proi)erties  of 
the  asymptotes  of  the  hyperbola. 


123.      The  tangents  (it  the  extremities  of  a  pair  of  conjugate 
diameters  intersect  on  the  asymptotes. 


Fig.  103. 

Let  Pi(xi,  Yi)  be  an  extremity  of  that  one  of  tlie  conjugate 
diameters  which  intersects  the  i)rimary  hyperbola.  The  tan- 
gent PiT  is 

Xix y^y 

3==  b^ 


1. 


242  PLANE    ANALYTIC    GKOMETRY. 

The  extremity  Qi  of  the  conjugate  diameter  is 


and  the  tangent  QiT  to  the  conjugate  hyperbola  is 
ayi     X    I    bxi     y  ^^ 
b      a'^        a      b" 

The  coordinates  of  the  point  of  intersection   T  of    these 
two  tangents  are  easily  found  to  he 

y  =  yi  +  -xi- 

a 

But  tliis  is  a  point  on  the  asymptote,  since  it  satisfies  the 
equation  y  =  -  x.  * 


124.  The  straight  line  connecting  the  extremities  of  a  pair 
of  conjugate  diameters  is  parallel  to  one  asymptote  and  bisected 
by  the  other. 

The  straight  line  connecting  the  points  Pi(xi,  yi)  and 
Qi[  -  yi,  -  Xi  )  (Fig.  103)  has  the  equation 

y  —  Yi      _    X  —  Xi 


b  a 

yi- 

which  reduces  to 


y.-^x,        x,-^y, 


y-yi  =  --  (x-x,). 

d. 

This  line  is  evidently  parallel  to  the  asymptote  y  = x. 

3. 

Moreover,  the  figure  CPiTQ,  is  a  parallelogram  by  §  121,  1, 
and  CT  is  an  asymptote  by  §  123.  Therefore,  QiPi  is  bisected 
by  CT,  since  the  diagonals  of  a  parallelogram  bisect  each 
other. 


THE    HYPERBOLA. 


243 


125.      The  portion  of  a  tangent   between  the  asi/mptotes  is 
bisected  by  the  point  of  tangency. 

For  in  Fig.  103,  PiT  =  CQi  and  PiT'  =  CQo,  being  opposite 
sides  of  parallelograms. 


But 

hence 


PiT=P,T'. 


126.      The  po)'tio?is  of  a  chord  between  the  hyperbola  and  its 
asymptotes  are  equal. 


Let  RQ  be  any  chord  intersecting  the  hyperbola  in  S  and 
M,  and  the  asymptotes  in  R  and  Q  ;  to  prove  QM  =  SR. 

Draw  the  diameter  CN  to  the  middle  point  of  SM,  and  at 
the  point  P  where  CN  intersects  the  hyperbola  draw  the  tan- 
gent TT'.  Then  by  §  121,  1,  TT'  is  parallel  to  SM.  But 
TP  =  PT',  by  previous  article, 

.-.  QN  =  NR. 
But  MN  =  NS,  by  construction, 

.-.  QM  =  SR. 


244 


PLANE    ANALYTIC    GEOMETRY. 


127.     Equation  of  an  Hyperbola  Referred  to  its  Asymp- 
totes as  Axes  of  Co  rdinates. 

We  wish  to  transform  the  equation  of  the  hyperbola 

^  _  y:  —  1 

a'      h'~    ' 
from  the  axes  CX  and  CY  to  the  axes  CX'  and  CY',  where 
CX'  and  CY'  are  the  asymptotes. 


Fig.  105. 

We  have,  by  [21],  §  45,  as  the  formulas  of  transformation 
X  ==  x'  cos  9  -\-  y'  cos  $', 
y  =  x'  sin  ^  +  y'  sin  6', 
where  d  =  XCX'  and  0'  =  XCY'. 

Since  CA  =  a  and  AH'  =  —  b, 

it  follows  from  Trigonometry  that 


cos  6 
Also,  since 


= ,  sin  0 


Va^+b^'  Va'+b^ 

CA  =  a  and  AH  =  b, 


cos  u'  = 


a  .     „,  b 

,  sin  6'  = 


Va^  +  b^ 


Va^  +  b^ 


THE    HYPERBOLA.  245 

Hence  the  formulas  of  transformation  become 
a 


=  (x'  +  y'), 


Va-  +  b^ 
•va-+  b- 


Substituting  in 
we  have 
or,  dropping  primes, 


a^      b^      -^' 


4x'y'     _ 


a-  +  b'  ^     -, 

xy==-^: —  [47] 


Similarly,  the  equation  of  the  conjugate  hyperbola  is 

a^+b^ 

^y= — 4— 

In  numerical  equations,  the  right-band  side  of  the  equation  [47]  will 
appear  as  a  single  number  and  we  shall  have  an  equation  of  the  form 

xy  =  c. 

To  obtain  the  axes  of  the  hyperbola,  reference  must  be  had  to  the 
angle  between  the  coordinate  axes.     For  example,  given  the  hyperbola 

xy  =  8, 

the  angle  between  the  coordinate  axes  being  G0°. 

Since  the  transverse  axis  of  the  hyperbola  bisects  the  angle  between 
the  coordinate  axes,  the  angle  between  the  transverse  axis  and  an  asymp- 
tote is  30°.     Hence, 

^  =  tan  :J0°  =  I  V3. 

Al.so  from  [47], 

a2  +  b2  _  ^ 

From  these  two  ecjuations  we  find 

a  =  2V0, 
b  =  2  V2. 


246  PLANE   ANALYTIC    GEOMETRY. 

128.     Polar  Equation  of  the  Hyperbola,  the  Pole  being  at 
the  Centre. 

To  obtain  the  polar  equation,  we  have   the    formulas    of 
transformation 

X  =  r  cos  ^,   y  =  r  sin  6. 
Substituting  in 


a^       b^"-"' 


we  obtain 


Whence, 


r2 

a^b^ 

b'-'  cos^  ^  -  a^ 

sin^  0' 

a^b^ 

(a^  +  b'O  cos-  e 

-a^' 

r=-  ..  ^; 

e'cos^e-1  W" 


y 


EXAMPLES. 

1.  Given  the  hyperbola  3x^  —  5y-  ^  7,  find  the  semi-axes, 
eccentricity,  foci,  directrices,  asymptotes,  and  latus  rectum, 
both  of  the  given  hyperbola  and  its  conjugate. 

2.  Find  the  equation  of  the  hyperbola  which  has  the  lines 
y  =  ±  fx  for  its  asymptotes  and  the  points  (±  4,  0)  for  its 
foci. 

3.  Find  the  equation  of  the  hyperbola  which  has  the  points 
(0,  zh  VT)  for  foci  and  passes  through  the  j^oint  (2,  —  1). 

4.  Find  the  equation  of  the  hyperbola  which  passes  through 
the  points  (—  2,  3)  and  (3,  —  5),  the  centre  of  the  curve  being 
at  the  origin  and  the  transverse  axis  coinciding  with  the  axis 
of  X. 

5.  Find  the  equation  of  the  hyperbola  with  eccentricity  3, 
which  passes  through  the  point  (2, 4),  the  centre  of  the  hyper- 
bola being  at  the  origin. 


THE    HYPERBOLA.  247 

6.  Find  the  equilateral    liyperbola  which  passes    through 
(5,  —  2),  the  centre  being  at  the  origin. 

7.  Find  the  equation  of  the  hyperbola,  the  vertices  of  which 
lie  halfway  between  the  centre  and  the  foci. 

8.  Does  the  point  (4,  5)  lie  inside  or  outside  the  hyperbola 
9x2  — 10y2  =  90? 

9.  Show  that  the  point  (xi,  yi)  lies  within,  on,  or  without 
the  hyperbola 

—,—  ,  :,  =  !,  according  as  ^  —  -j^  —  1  >,  =  ,  or  <  0. 
a-      b-  a^       b- 

10.  If  e  and  e'  are  the  eccentricities  of  the  primary  and 
the  conjugate  hyperbolas,  respectively,  show  that  —^-] — ;;;  =  1- 

11.  If  the  vertex  lies  two  thirds  of  the  distance  from  the 
centre  to  the  focus,  find  the  slopes  of  the  asymptotes. 

12.  Express  the  angle  between  the  asymptotes  in  terms  of 
the  eccentricity  of  the  hyperbola. 

13.  Show  that  the  eccentricity  of  an  equilateral  hyperbola 
is  the  ratio  of  the  diagonal  of  a  square  to  its  side. 

14.  Show  that  in  an  equilateral  hyperbola  the  distance  of  a 
point  from  the  centre  is  a  mean  proportional  between  its  focal 
distances. 

15.  Find  the  equations  of  the  tangent  and  the  normal  to 
the  hyperbola  9x^  —  4y^  =  36  at  a  point  the  abscissa  of  which 
is  equal  to  its  ordinate. 

X"  V" 

16.  Find  the  tangents  to  the  hyperbola  q  "~  ^  =  ~  ^  which 

are  parallel  to  the  straight  line  5y  —  2x  +  3  =  0. 

17.  Find  the  lengths  of  the  subtangent  and  the  subnormal 
for  the  point  (3,  2)  on  the  hyperbola  x^  — 2y-  =  l;  also  the 
lengths  of  the  portions  of  the  tangent  and  the  normal  in- 
cluded between  the  curve  and  the  transverse  axis. 

18.  Show  that  the  subtangent  and  tlie  subnormal  for  the 

point  (xj,  Yi)  on  the  hyperbola 

x2      y2       .  ...      x,'-a'       ,  b^ 

—  —  —,  =  1  are,  respectively,  — ■ and  —  Xi. 

a'       b"  Xi  a 


248  PLANE    ANALYTIC    GEOMETRY. 

19.  Find  where  the  tangents  from  the  foot  of  the  directrix 
will  meet  the  hyperbola,  and  what  angles  they  will  make 
with  the  transverse  axis. 

20.  Prove  that  tangents  to  the  conjugate  hyperbola  at  the 
points  where  it  is  cut  by  the  tangent  at  the  vertex  A  of  the 
primary  hyperbola  pass  through  the  other  vertex  A'. 

21.  If  any  number  of  hyperbolas  have  the  same  transverse 
axis,  show  that  tangents  to  the  hyperbolas  at  points  having 
the  same  abscissa  all  pass  through  the  same  point  on  the 
transverse  axis. 

22.  Prove  that  an  ellipse  and  hyperbola  with  the  same  foci 
cut  each  other  at  right  angles. 

23.  Show  that  the  portion  of  the  normal  between  the  axes 
is  divided  by  the  curve  in  the  ratio  a^  :  bl 

24.  If  the  tangent  at  any  point  P  on  the  hyperbola  inter- 
sects the  transverse  axis  at  T,  and  if  CP  meets  the  tangent  at 
the  vertex  A  at  R,  show  that  RT  is  parallel  to  A  P. 

25.  If  a  tangent  to  an  hyperbola  is  intersected  by  the  tan- 
gents at  the  vertices  in  the  points  Q  and  R,  show  that  the  circle 
described  on  QR  as  a  diameter  passes  through  the  foci. 

26.  In  the  hyperbola  4x"  —  2y^  =  7,  find  the  equation  of 
the  conjugate  diameters,  one  of  which  bisects  the  chord 
3x  -  2y  +  3  =  0. 

27.  In  the  hj^perbola  x-  —  3y^=  6,  find  the  equation  of  the 
chord  the  middle  point  of  which  is  (4,  1). 

28.  Given  the  point  (i,  |)  on  the  hyperbola  48x'-'  — 4y^  =  3, 
find  the  point  on  the  conjugate  hyperbola  at  the  extremity  of 
the  corresponding  conjugate  diameter. 

29.  Find  the  equation  of  the  hyperbola  4x^  —  9y'-'  =  36 
referred  to  a  pair  of  conjugate  diameters  as  axes,  one  of 
which  makes  an  angle  30°  with  the  transverse  axis. 

30.  Show  that  every  diameter  of  an  equilateral  hyperbola 
is  equal  to  its  conjugate. 

31.  Sho\^that  in  an  equilateral  hyperbola  conjugate  diam- 
eters are  equally  inclined  to  the  asymptotes. 


THE    HYPERBOLA.  249 

32.  Show  that  in  an  equilateral  hyperbola  all  diameters  at 
right  angles  to  each  other  are  equal. 

33.  Show  that  the  polar  of  any  point  on  a  diameter  is  par- 
allel to  the  conjugate  diameter. 

34.  Show  that  if  an  ellipse  and  an  hyperbola  have  the 
same  axes  in  magnitude  and  position,  then  the  asymptotes 
of  the  hyperbola  coincide  with  tlie  equal  conjugate  diameters 
of  the  ellipse. 

35.  Show  that  the  area  of  the  triangle  made  by  any  tangent 
and  its  intercepts  on  the  asymptotes  is  constant. 

36.  Show  that  the  focal  distance  of  any  point  on  the 
hyperbola  is  equal  to  the  line  drawn  through  the  point  par- 
allel to  an  asymptote  to  meet  the  directrix. 

37.  A  perpendicular  is  drawn  from  a  focus  of  an  hyperbola 
to  an  asymptote  ;  show  that  its  foot  is  at  a  distance  a  and  b 
from  tlie  centre  and  the  focus,  respectively. 

38.  Show  how  an  hyperbola  may  be  graphically  constructed 
by  aid  of  §  126,  if  the  asymptotes  and  one  point  of  the  curve 
are  given. 

39.  If  a  circle  be  described  from  the  focus  of  an  hyperbola 
with  radius  equal  to  half  the  conjugate  axis,  it  will  touch  the 
asymptotes  in  the  points  where  they  intersect  the  directrix. 

40.  If,  from  any  point  P  of  an  hyperbola,  PK  is  drawn 
parallel  to  the  transverse  axis,  cutting  the  asymptotes  in  Q 
and  R,  then  PQ  .  PR^a";  if  PK  is  drawn  parallel  to  the 
conjugate  axis,  then  PQ  .  PR  :=  —  b^. 

41.  If  two  concentric  equilateral  hyperbolas  be  described, 
the  axes  of  one  being  the  asymptotes  of  the  other,  show  that 
tliey  will  intersect  at  right  angles. 

42.  Find  the  equation  of  2x^  —  3y^  =  6  referred  to  its 
asymptotes  as  coordinate  axes. 

43.  Find  the  semi-axes,  eccentricity,  and  vertices  of  the 
hyperbola  xy  =  1,  the  axes  being  rectangular. 

44.  Find  the  semi-axes,  eccentricity,  and  vertices  of  the 
hyperbola  xy  ^  —  12,  tlie  angle  between  the  axes  being  60°. 


250  PLANE    ANALYTIC    GEOMETRY. 

45.  Show  that  xy  +  ax  +  by  +  c  =  0  is  the  general  equa- 
tion of  the  hyperbola,  when  the  axes  of  coordinates  are 
parallel  to  the  asymptotes. 

46.  Prove  that  the  equation  of  the  tangent  to  the  hyperbola, 

referred   to   its    asymptotes    as  axes,  is  Xiy  +  yiX  =  — ~ — , 

where  (xi,  yj)  is  the  point  of  contact. 

47.  If  the  ordinate  N  P  of  an  hyperbola  be  produced  to  Q, 
so  that  NQ  =  FP,  find  tlie  locus  of  Q. 

48.  AOB  and  COD  are  two  straight  liiaes  which  bisect  each 
other  at  right  angles;  find  the  locus  of  a  point  which  moves 
so  that  PA.  PB=  PC.  PD. 

49.  A  straight  line  has  its  extremities  on  two  fi.xed  straight 
lines  and  makes  with  them  a  triangle  of  constant  area ;  find 
the  locus  of  the  middle  point  of  the  line. 

50.  Given  a  fixed  line  AB  and  a  fixed  point  Q;  from  any 
point  R  in  AB  a  perpendicular  is  drawn  equal  in  length  to 
RQ ;  find  the  locus  of  the  extremity  of  the  perpendicular. 

51.  A  chord  PQ  of  an  ellipse  is  perpendicular  to  the  major 
axis ;  PA  and  QA'  are  produced  to  meet  in  R  ;  show  that  the 
locus  of  R  is  an  hyperbola  having  the  same  axes  as  the  ellipse. 

52.  Show  that  the  locus  of  the  centres  of  all  circles  which 
touch  two  given  circles  is  an  hyperbola  or  an  ellipse. 

53.  Find  the  locus  of  the  intersection  of  a  pair  of  perpen- 
dicular tangents  to  an  hyperbola. 

54.  Find  the  locus  of  the  foot  of  tlie  perpendicular  from 
either  focus  of  an  hyperbola  to  any  tangent. 


CHAPTER   XI. 

THE    GENERAL    EQUATION    OF    THE    SECOND    DEGREE. 

129.     Statement  of  Problem. 

The  most  general  form  of  the  equation  of  the  second 
degree  is 

Ax-+2  Hxy+ By2  +  2  Gx  +  1>  Fy  +  C=0,  (1) 

where  the  coefficients  A,  H,  B,  G,  F,  C  may  have  any  values, 
positive  or  negative  or  zero,  except  that  all  three  of  the 
coefficients  A,  H,  B  shall  not  be  zero  at  the  same  time.  For, 
in  that  case,  the  equation  would  not  be  of  the  second  but 
of  the  first  degree.  This  equation  evidently  includes  that 
treated  in  chapter  VI,  §  70,  for  the  latter  is  obtained  from 
(1)  by  placing  H  ^  0. 

We  propose  now  to  examine  the  general  equation,  and  to 
show  that,  whatever  may  be  the  values  of  the  coefficients,  the 
locus  represented  is  one  of  the  conic  sections,  under  which 
name  we  include  not  only  the  circle,  ellipse,  hyperbola,  and 
parabola,  but  also  the  limiting  cases  noticed  in  chapter  VI. 

The  method  to  be  followed  is  to  reduce  the  equation  to  its 
simplest  form  by  a  proper  choice  of  new  axes  of  coordinates. 
It  will  be  found  that  the  final  form  obtained  depends  upon 
certain  hypotheses  respecting  the  values  of  the  coefficients, 
but  that  the  possible  results  are  limited  to  just  those  equations 
which  we  know  represent  conic  sections. 

To  simplify  the  Avork  we  shall  assume  for  the  present  that 
the  equation  is  referred  to  rectangular  axes.  The  case  of 
oblique  axes  will  be  liandled  briefly  in  §  136. 


252  PLANE    ANALYTIC    GEOMETRY. 

The  general  equation  has  already  been  partially  handled  in 
chapter  III,  §  41,  and  it  was  shown  that,  if 

ABC  +  2  FGH- AF^-BG--CH2  =  0, 

the  equation  represents  two  straight  lines.  This  fact  will  be 
proved  anew  in  the  course  of  our  present  work.  Therefore 
we  do  not  need  to  assume  it,  though  it  is  well  to  bear  it  in 
mind. 

130.     To  Remove  Terms  of  the  First  Degree. 

We  first  proceed  to  see  if  it  is  possible  to  reduce  the  equa- 
tion so  that  the  terms  of  the  first  degree  in  x  and  y  shall  dis- 
appear. For  that  purpose,  let  us  move  the  axes  parallel  to 
themselves,  by  placing 

X  —  Xq  -j-  X  ,  ,„, 

y  =  yo  +  y',  ^^ 

Xo  and  Yo,  the  coordinates  of  the  new  origin,  being  left  unde- 
termined for  a  moment. 

Substituting  these  quantities  in  (1),  we  have 

Ax'2  -f  2  HxV  +  By'2  -f  2(Axo  +  Hyo  +  G)  x'  +  2(Hx,  +  Byo 
+  F)y'+Ax,^  +  2  Hxoyo+By„^  +  2  Gxo  +  2  Fyo+C  =  0. 

If  now  we  can  choose  for  Xq  and  yo  such  quantities  that  the 
coefficients  of  x'  and  y'  shall  vanish,  we  shall  accomplish  our 
object.     We  must  have 

Axo+Hyo+G  =  0, 

Hxo+Byo+F  =  0.  ^^ 

The  solution  of  these  equations  is  possible,  when  AB  —  H'- 
is  not  zero,  giving  as  the  necessary  values  of  Xq  and  yo, 

_  HP-  BG 
'°"AB-H^   '  M) 

_HG- AF  '^  ^ 

y"~AB-H'-^' 


THE    GENERAL   EQUATION.  253 

If  AB —  H^  =  0,  however,  the  equations  (3)  are  contradic- 
tory, and  hence  no  vahies  of  Xq  and  yo  can  be  found,  wliicli 
will  make  the  coefficients  of  x'  and  y'  vanish. 

There  is  one  case  even  when  AB  —  H-  =  0,  in  which  values  of  xq  and 
yo  can  be  found.     This  is  the  case  in  which  the  relations  hold 

H~  B        F  ■ 

The  equations  (3)  are  then  identical,  and  values  of  xo  and  yo  which 
satisfy  either  will  make  the  coefficients  of  x'  and  y'  vanish.  This  case 
will  be  fully  provided  for  later,  and  need  not  be  discussed  here. 

We  will  distinguish  three  cases,  in  the  following  discussion, 
namely  : 

I,  AB-  H->0;  II,  AB-  H2<0;  III,  AB  -  H'^^O. 

But  before  passing  to  them  we  will  sum  up  the  results  of 
this  paragraph  as  follows  : 

The  terms  of  the  first  degree  may  he  removed  from  the  general 
equation  jnovided  the  quantity  AB  —  W  is  not  zero.  The  equa- 
tion theyi  talces  the  form 

Ax'2+2Hx'y'+By'^  +  (Axo^  +  2Hx,y,+  Byo^+2Gxo  +  2Fyo 

+  C)=0, 

whence  it  is  seen  that  the  coefficients  of  the  terms  of  the  second 
degree  are  unchanged,  and  the  absolute  term  is  ■  obtained  by 
substituting  the  coordinates  of  the  neiv  origin  in  the  original 
equation. 

131.     Case  I.    AB  -  H-  >  0. 

We  have  just  seen  that  by  means  of  a  cliange  of  axes  the 
equation  may  be  reduced  to 

Ax'2  +  2  Hx'y'  +  By'-  +  C  ==  0,  (5) 

where 

C  =  Ax,r  +  2  Hx,y,  +  By,^  +  2  Gx^  +  2  Fy,  +  C. 


254  PLANE   ANALYTIC    GEOMETRY. 

The  value  of  C  may  be  expressed  in  terms  of  the  original 
coefficients  as  follows. 

Take  the  two  equations  (3),  multiply  the  first  one  by  Xq,  the 
second  by  yo,  and  add  them.     There  results 

Axo^  +  2  HxoYo  +  Byo^  +  Gxo  +  Fyo  =  0. 
Subtracting  this  sum  from  the  value  of  C,  we  have 
C'=Gxo+Fyo+C, 
whence,  finally,  by  substituting  the  values  of  Xq  and  yo,  as 
given  in  (4),  we  have 

^,      ABC  +  2  FGH-AF2-BG2-CH- 
^^ AB^^hP ^^> 

The  quantity  in  the  numerator  is  called  the  Discriminant 
of  the  equation  (1)  and  is  usually  denoted  by  the  Greek  letter 
A.  If  we  represent  the  quantity  AB  —  H^  by  D,  we  have 
then 

Returning  now  to  the  equation  (5), 

Ax'2  +  2  Hx'y'+By'2+C'  =  0, 
we  will  next  endeavor  to  remove  the  term  2  Hx'y'.  For  that 
purpose  we  turn  the  axes  through  an  angle  $,  keeping  the 
origin  fixed  and  the  axes  rectangular.  The  three  sets  of 
axes  we  have  now  used  lie  as  shown  in  Fig.  106,  the  figure 
being  drawn  on  the  assumption  that  Xo,  yo?  and  6  are  all  posi- 
tive. 

The  formulas  of  transformation  to   the  new  axes  are  by 

[20],  §44, 

x'  =  x"  cos  0  —  y"  sin  6, 
y'  =  x"  sin  6  -\-  y"  cos  6. 

Substituting  in  (5),  we  have 

(A  cos^  ^  +  2  H  sin  ^  cos  e  +  B  sin^  6)  x'"' 
+  2  {(B  —  A)  cos  6  sin  ^  +  H  (cos^  $  -  sin^  6)}  x"y" 
+  (A  sin^  ^  -  2  H  sin  ^  cos  e  +  B  cos^  6)  y"-  +  C  =  0. 


THE   GENERAL   EQUATION. 


255 


Fig.  106. 


We  must  now  determine  6  so  that  the  coefficient  of  x"y" 
shall  vanish;  i.e.,  so  that 

2  (B  —  A)  cos  ^  sin  ^  +  2  H  (cos^  d  —  sin^  0)  =  0. 
This  equation  is  equivalent  to 

2  H  cos  2  ^  +  (B  —  A)  sin  2^  =  0, 
2  H 


whence 


or 


tan  2  e 


=  4  tan" 


-  B' 
2  H 


A-B  (») 

Among  the  various  values  which  6  may  have  we  shall 
agree  always  to  choose  that  one  which  makes  2  6  lie  between 
0°  and  180°.     Hence  9  is  always  a  positive  acute  angle. 

With  this  value  of  9,  our  equation  takes  the  form 

A'x"-+ B'y'"+C'==0,  (9) 

where        A'  =  A  cos^  9 -\- 2  H  sin  9  cos.  9  -\- B  sin^  9, 
and  B'  =  A  sin^  ^  —  2  H  sin  ^  cos  ^  +  B  cos"  9. 

These  values  of  A'  and  B'  may  be  expressed  in  terms  of  the 
original  coefficients  by  use  of  (8). 


256  PLANE    ANALYTIC    GEOMETRY. 

We  have 

A'  1=  A  cos^  6  +  2  H  siu  ^  cos  ^  +  B  siu^  0, 

.l±cos20.^    .     ^  ^  p  1  -  cos  2  g 
=  A 2 h  H  sm  2  ^  B , 

=  i  [A  +  B  +  (A  -  B)  cos  2  6  +  2  H  sin  2  ^]. 

But  since,  by  (8), 

2  H 
tan  2  0  = 


A-  B' 
it  follows  that 

■    on  2  H  ;        „-  A-B 

sm  2  ^  =  -  and  cos  2  6  = 


V(A-B)-  +  4H2  V(A-B)^  +  4H^ 

Hence 

A-  =  irA+B+     (A-B)-  +  4Hn 


or  A'  =  i[A  + B  +  V(A-B)2  +  4  H^J. 

Similarly, 

B'  =  |(A  +  B  -  V(A-  B)'^  +  4  H^. 

From  these  values  we  obtain  the  following  relations  : 

A'+B'=A+B, 
A'B'=  AB-  H^ 

2  H 


(10) 


A'-B'  =  V(A-  B)^  +  4  H2  =  -^^— . 
^  ^  sin  2  6 


(11) 


J 


The  first  two  of  these  relations  give  sufficient  data  to  com- 
pute A'  and  B'  from  A  and  B  ;  any  possible  ambiguity  in  the 
result  being  removed  by  the  last  equation,  which  shows  that 
A'  —  B'  has  the  same  sign  as  H,  since  sin  2  ^  is  always  positive. 

Moreover,  since,  in  the  case  before  us,  AB  —  H^  is  positive, 
it  follows  from  the  second  relation  that  A'  and  B'  have  the 
same  sign. 

We  are  now  ready  to  see  what  locus  is  represented  by  the 
equation  (5).     An  important  difference  arises  according  as 


THE    GENERAL    EQUATKJN.  257 

C  is  or  is  not  zero,  or,  what  is  the  same  thing  [  since  C  =  ^ 
according  as  A  is  or  is  not  zero. 

(1)  A  <  0. 

We  may  then  write  the  equation  in  the  form 
A'x"-+B'y"'  =  -C' 

A'  B' 

Since  A'  and  B'  have  the  same  sign,  the  denominators  must 
be  either  both  positive  or  both  negative.  In  the  first  case,  the 
equation  may  be  written 

a-^        b- 
and  the  locus  is  an  ellipse,  including  the  special  case  of  a 
circle. 

In  the  second  case,  the  equation  may  be  written 
in  ,12 

a-       b 
and  the  locus  is  an  "  imaginary  ellipse." 

(2)  A  =  0. 

The  equation  then  is 

A'x"-+  B'y'"  =  0. 

A'  and  B',  having  the  same  sign,  can  both  be  made  positive, 
and  the  equation  may  be  written 

and  the  locus  is  ix  point. 

Summing  up  briefly,  we  say  : 


258  PLANE    ANALYTIC    GEOMETRY. 

If  AB —  H">0,  the  general  equation  of  the  second  degree 
may  he  reduced  to  one  of  the  three  following  forms  : 

- — I-  ^-  =  1 
a^  ^  h""         ' 

a  b 

a-        b" 

The  locus  represented  is  a  real  elUptse  (including  the  circle^, 
an  imagiiiart/  ellipse,  or  a  point 

It  was  proved  in  chapter  III  that  if  A  =  0,  the  general  equation  rep- 
resented two  straiglit  lines.  Here  we  have  a  case  where  A  =  0,  and  the 
equation  represents  a  point.  The  contradiction  is  only  apparent.  For 
we  may  factor 


into 


(7  +  ^'^^)(7-f^-)  =  »- 


Each  factor  placed  equal  to  zero  gives  an  equation  of  the  first  degree 
and  hence  may  be  said  to  represent  a  straight  line,  but  because  of  the 
V—  1  involved,  no  real  point  lies  on  either  line  except  the  point  (0,  0). 
We  may  say  either  that  the  equation  represents  "  two  imaginary  straight 
lines,"  or,  if  we  confine  ourselves  to  real  quantities,  that  it  represents  a 
point. 


132.     Case  II.    AB  -  H"  <  0. 

The  work  proceeds  exactly  as  in  Case  I,  and  we  find,  as 
before,  the  simple  equation 

A'x"-+ B'y"'+C'  =  0, 

where   A'.    B',  and  C  are  defined  by  the  same  formulas  as 
before. 


THE    GENERAL   EQUATION.  259 

But  here  A'  and  B'  have  opposite  signs,  since 

A'B'  =  AB  -  H-,  and  AB  -  H'  <  0. 

As  before,  we  consider  two  cases. 

(1)  A<0. 

We  may  then  write  the  equation  in  the  form 


which  is  either 


C'   '         C 
A' 


or  ^+^z=rl. 

The  locus  in  either  case  is  an  hi/jjerhola. 

(2)  A  =  0. 

Tlie  equation  then  is 

A'x"-+  B'y"2  =  0, 
which  may  be  written 

a-        b-^ 
Hence  the  locus  consists  of  two  intersecting  straight  lines. 
Summing  up,  we  say  : 

If  AB —  H'- <  0,  the  general  equation  of  the  second  degree 
may  be  reduced  to  one  of  the  three  following  forms : 

^'  _  y!!  —  1 

a^        b^  ~    ' 
a^  "^  b^  ~    ' 

a-        b- 

The  locus  represented  is  either  an  hyperbola  or  two  intersect- 
ing straight  lines. 


260  PLANE    ANALYTIC    GEOMETRY. 

133.     Case  III.    AB  —  H^  =^  0. 

As  we  have  seen  in  §  130  it  is  not  possible  to  begin,  as  in 
Cases  I  and  II,  by  removing  the  terms  of  the  first  degree. 
We  must  try  some  other  method  of  procedure,  and  the  follow- 
ing seems  most  convenient. 

Since,  by  hypothesis,  AB  —  H^  =  0,  the  terms  of  the  second 
degree  in  the  general  equation  make  a  perfect  square,  and  the 
equation  may  be  written 

(VAx+ VBy)2  +  2  Gx  +  2  Fy  +  C  =  0,  (12) 

where  VB  is  taken  with  the  same  sign  as   H. 

We  will  now  assume  such  axes,  OX'  and  OY',  that 

VAx  +  VBy  =  0 

shall  be  the  equation  of  OX'  with  respect  to  the  original  axes. 
The  line  perpendicular  to  OX',  namely 

VBx  —  V  Ay  =  0, 

will  then  be  OY'.    Then,  since  x'  and  y'  are  the  perpendiculars 
on  OY'  and  OX'  respectively, 

,       VBx  —  VAy 

x'  = ,  } 

VA+B 

VAx  +  VBy 
y  =^ — • 

Va  +  b 

Solving  these  equations  for  x  and  y,  we  have 
_  VBx'  +  VAy^ 

-^1+^    I  (13) 

_-VAx'  +VBy' 

^~       Va+b 

A  comparison  of  these  formulas  with  [20],  §  44,  shows  that 

Vb 

OX'  makes  an  angle  6  with  OX,  where  cos  $=^    , — =  and 

Va+b 


THE    GENERAL   EQUATION.  261 

-  Va 

sin  B  =    ,  .     Hence  ^  is  a  neaatiue  anale,  and  is  acute 

VA+  B 

or  obtuse  according  as  VB  is  positive  or  negative. 
Substituting  the  values  of  x  and  y  in  (12),  v^  have 

VA+  B  VA+  B 

or  y'-'  +  2  G'x'  +  2  F'y'  +  C  ==  0,  (14) 

-,     gVb-fVa       GH-FA 

where  G  — 

F'  = 


V(A-(-B)3  VA(A+B)3 

G  V A  +  F  Vb  _    GA+  FH 
V(A+B)3     ~  V A(A  +  Ef 
C 


A+  B 


The  partial  rationalization  of  G'  and  F'  is  effected  by  multiplying 
numerator  and  denominator  by  Va. 

This,  of  course,  would  not  be  possible  if  A  =  0 ;  but  in  that  case 
H  =  0  (since  AB  =  H^),  and  hence  the  equation  is  already  in  the  form 
we  are  seeking  to  obtain,  and  may  be  treated  as  in  §  70. 

The  further  classification  now  depends  upon  the  question 
whether  the  value  of  G'  be  zero  or  not.  But  this  in  turn 
depends  upon  A.  For  if  we  substitute  H  =  VAB  in  the  ex- 
pression for  A  given  in  §  131,  we  find 

A  =  2  FG  VAB-AF^-BG^ 

=  -  (G  Vb  -  F  VA)-  =  -  (A  +  B)'  G'-. 

Hence,  if  A  <  0,  G'  <  0,  and  if  A  =  0,  G'  =  0. 
We  examine  each  case  separately. 

(1)  A  <  0. 

Then  G'  is  not  zero,  and  the  equation  (14)  contains  a  term 
in  x'.  The  equation  may  be  handled,  therefore,  by  tlie  method 
of  completing  the  square  as  in  chapter  VI,  §  70. 


262  PLANE   ANALYTIC   GEOMETRY. 

There  results  finally 

y"2  =  4px", 
which  is  the  equation  of  a  parabola. 
(2)  A  =  0. 

Then  G'  =  0,  and  the  equation  (14)  contains  no  term  in  x', 
but  is  simply 

y'2  +  2  F'y'+C'  =  0. 

By  the  theory  of  quadratic  equations,  this  equation  may  be 

factored  into 

(y'-a)(y'-^)=0, 

i 
where  a  and  /3  are  the  roots  of  the  equation. 

Hence  the  locus  consists  of  two  parallel  straight  lines, 
ivhich  7nay  be  real,  coincident,  or  imaginary. 

Summing  up,  we  say  : 

If  AB —  H'^^^O,  the  general  equation  of  the  second  degree 
may  be  reduced  to  one  of  the  two  follounng  forms : 

y"2  =  4px", 
(y'-a)(y'-^)-0. 

The  locus  represented  is,  therefore,  either  a  pjarahola  or 
two  parallel  straight  lines,  which  may  be  real,  coincident,  or 
imaginary. 

134.     Summary. 

The  results  of  the  previous  articles  are  exhibited  in  the 
following  table  which  gives  the  simplest  form  to  which  the 
general  equation  may  be  reduced  under  the  various  hypoth- 
eses. 

General  Equation. 

Ax"  +  2  Hxy  +  By2  +  2  Gx  +  2  Fy  +  C  =  0. 

A  =  ABC +  2  FGH-  AF2-BG--CH^ 

D  =  AB-  H^. 


THE    GENERAL    EQUATION. 


263 


A^o 

A  =  o 

D>0 

Ellipse 

¥''''2         m'^'I 

V  +  v  =  1 

a-         b- 
Imaginary  Ellipse 

z?        b2 

Point 

(Two  imaginary  inter- 
secting straight  lines) 

a^         b- 

D  <0 

Hyperbola 
a-         b- 

Two  intersecting 
straight  lines 

a-         h- 

D  =  0 

Pai-abola 
y"2  =  4px" 

Two  parallel  straight 

lines  (real,  coincident, 

or  imaginary) 

(y'  -  a)(y'  -  iS)  =  0 

135.     Rule  for  Handling  Numerical  Equations. 

Out  of  the  foregoing  discussion  we  may  deduce  tlie  follow- 
ing working  rule  for  the  reduction  of  any  given  numerical 
equation. 

Compute  AB  —  H^. 

I.    7/'AB-H2<0. 

1.   Substitute 

X  =  Xo  +  x', 

y  =  Yo  +  y', 

and  determine  Xq  and  yo  so  that  the  coefficients  of  x'  ami  y'  shall 
vanish.  Compute  the  absolute  term  C,  by  s^chstitutiny  Xq  and  yo 
in  the  given  equation. 


264  PLANE   ANALYTIC   GEOMETRY. 

2.  Compute  A'  and  E'  from  the  relations 

A'+  B'=A+  B, 
A'B'=  AB-H^, 

noting  that  A'  —  B'  has  the  same  sign  as  H. 

3.  The  equation  is  noiv 

A'x"-+  B'y"'+C'=0, 
and  the  form  of  the  locus  is  at  once  evident. 

4.  The  position  of  O'X"  is  found  by  passing  a  line  through 

2  H 

(xo,  yo)  making  the  acute  angle  ^  tan"' ~  with  OX. 

A  —  B 

II.   If  AB-  H2  =  0. 

1.  Write  the  equation  in  the  form 

(  VAx  +  VBy)'  +  2  Gx  -f  2  Fy  +  C  =  0, 
VB  being  taken  ivith  the  same  sign  as  H, 

and.  let  y'  ^ , -? 

Va  +  b 

VBx  —  VAv 

X  ^ — —  • 

Va  +  b 

2.  Solve  these  equations  for  x  cmd  y  in  terms  of  x'  and  y'  and 
substitute  in  the  given  equation. 

3.  The  equation  is  now  in  the  form 

y'2  +  2  G'x'  +  2  F'y'  +  C'  =  0, 
and  the  further  reduction  is  to  be  made  by  completing  the  square. 

4.  The  p)ositio7i  of  OX'  is  found  by  turning  OX  through  a 

negative  angle,  tan~^  (  — -=^ 

Ex    1 

8x2  -  4xy  +  5y2  _  36x  +  18y  +  9  =  0. 

Here,  A  =  8,     B  =  5,     H  =  —  2. 

...  AB-  H-^  =  3G. 


THE    GENERAL    EQUATION. 


265 


(1)  We  place  x  =  xo  +  x', 

y  =  yo  +  y', 

obtaining 

8x'2  -  4x'y'  +  5y'2  +  (16xo  -  4yo  -  3G)  x'  +  (-  4xo  +  lOyo  +  18)  y'  +  Sxq^ 
-  4xoyo  +  5yo-  -  36x0  +  18yo  +9  =  0. 
If  we  place  16xo  —  4yo  —  36  =  0, 

-4xo+10yo+18  =  0, 
we  find  xo  =  2,        yo  =  ~  1- 

Hence  the  reduced  equation  is 

8x'2  -  4x'y'  +  5y'2  -  36  =  0. 

(2)  A'  +  B'  =  13, 

A'B'  =  36. 

From  these  equations  follow 

A'  =  4  or  9,         B'  =  9  or  4. 

But  since  H  is  negative,  we  take 

A'  =  4,         B'  =  9. 

(3)  The  equation  is  now 

4x"-  +  9y"2  =  36. 

Y 


Fig.  107. 


266  PLANE   ANALYTIC    GEOMETRY. 

This  is  an  ellipse,  of  which  the  major  axis  along  O'X"  is  6,  and  the 
minor  axis  is  4. 

(4)  The  line  O'X"  passes  through  (2,  —  1)  and  makes  angle 
i  tan-i  (—  f)  =  tan-i  2  with  OX. 

Ex.2. 

3Gx2  -  48xy  +  lGy2  +  52x  -  200y  -  39  =  0. 

A  =  36,    B  =  16,    H  =  -  24. 

.-.  AB-  H2  =  0. 


(1)  We  write 
and  place 


(2)  Then 


(6x  -  4y)2  +  52x  -  260y  -  39  =  0, 

,  _  6x  —  4y  _  3x  —  2y 

^  V52     ~      Vis    '  - 

,      —  4x  — 6y       —  2x  —  3y 

X    ^   ^rz — -   ^  = ^  * 

V52  Vl3 


-  2x'  +  3y' 

Vl3 

—  3x'  —  2y' 


Substituting,  we  have,  after  reduction, 

~4 


y'2  +  Vl3y'  +  Vl3x'  -  ?  =  0. 


(3)  Completing  the  square,      * 

(,.f7=-Vir=(.-4), 

or  finally,  y"^  =  —  Vl3x", 

,,_    ,  ,  Vl3     ,,_    ,         4 

where  y    —  Y  +—?;—?  x    =  x ^^• 

2  Vl3 

(4)  The  axis  of  the  parabola  is  parallel  to  OX',  the  equation  of  which 
is  3x  -  2y  =  0. 

The  positive  extremities  of  the  axes  are  marked  in  Fig.  108. 


THE   GENERAL   EQUATION. 


267 


Y 

. 

/ 

/ 

/ 

A 

\S"          ^\ 

^ 

l^ 

V 

/ 

/  x^ — ^ 

-K, 

{" 

X' 

Xy" 

\v' 

Fig.  108. 


136.     Oblique  Coordinates. 

We  have  assumed,  thus  far,  that  the  general  equation  is 
referred  to  rectangular  coordinates.  If,  however,  the  equa- 
tion 

Ax*-^  +  2  Hxy  +  By"  +  2  Gx  +  2  Fy  +  C  =  0 

has  reference  to  oblique  coordinates,  it  may  be  transformed 
to  any  conveniently  chosen  pair  of  rectangular  coordinates. 
Formulas  for  this  purpose  are  given  in  §  46,  and  it  has  been 
proved  in  §  48  that  su.ch  a  transformation  does  not  alter  the 
degree  of  the  equation.  Therefore,  the  new  equation  is  of 
the  form 

A'x'-^  +  2  H'x'y'  +  B'y"  +  2  G'x'  +  2  F'y'  +  C  =  0. 


268  PLANE  ANALYTIC  GEOMETRY. 

This  equation  may  now  be  investigated  by  the  methods  of 
this  chapter. 

Hence  we  have  the  result : 

Any  equation  of  the  second  degree,  whether  referred  to  rect- 
angular or  to  oblique  coordinates,  re2)resents  one  of  the  conic 
sections. 

137.     Conic  through  Given  Points. 

The  general  equation  of  the  conic  section, 

Ax^  +  2  Hxy  +  By2  +  2  Gx  +  2  Fy  +  C  =  0, 

contains  six  coefficients.  However,  only  five  of  these  are 
essential,  since  we  may  divide  the  equation  by  any  one  of  them 
without  altering  its  meaning.  For  example,  we  may  make 
the  last  coefficient  C  equal  to  1,  except  in  the  case  when  the 
conic  passes  through  the  origin.  It  follows,  therefore,  that 
in  general  one  and  only  one  conic  may  he  2)assed  thi'ongh  any 
five  points  in  the  plane.  For,  if  we  substitute  for  x  and  y  the 
coordinates  of  the  given  points,  we  shall  have  five  linear 
equations  out  of  which  to  find  the  values  of  the  five  essential 
constants.  A  more  convenient  method  of  procedure  is  as 
follows  : 

Let  us  take  any  four  of  the  given  points  and  connect  them 
by  straight  lines,  so  as  to  form  a  quadrilateral  (Fig.  109). 

Let  the  equation  of  PiPg  be  Ajx  +  Bjy  +  Ci  =  0,  or,  more 
shortly,  Ui  =  0,  where  Ui  is  simply  an  abbreviation  for  the  left- 
hand  member  of  the  equation.  Similarly,  let  the  equation 
of  P2P3  be  U2  =  0,  that  of  P3P4  be  U3  =  0,  and  that  of  P4P1  be 

Form  now  the  equation 

UiU3  +  kU,U4  =  0,  (1) 

where  k  is  an  undertermined  factor.  This  equation  is  of  the 
second  degree  in  x  and  y ;  therefore,  it  represents  a  conic 
section.     Moreover,  this  conic  section  passes  through  P^ ;  for 


THE   GENERAL   EQUATION.  269 


Fig.  lOD. 

the  coordinates  of  Pi  make  Ui  =  0  and  U4  =  0,  and  therefore 
satisfy  equation  (1).  Similarly,  this  conic  passes  through 
Po,  P3,  and  P4.  If  now  we  substitute  in  (1)  the  coordinates 
of  P5,  we  determine  the  value  of  k  which  we  must  assume  in 
order  that  the  conic  may  pass  through  P5.  We  thus  deter- 
mine the  equation  of  a  conic  through  the  five  given  points. 

Ex.     Let  it  be  required  to  pass  a  conic  tlirougli  tlie  points 

Pi (2,  3),  P2(-l,  2),  P3(-3,  -1),  P4(0,  -4),  P5(l,  1). 

The  equation  of  PiPo  is  x  —  3y  +  7  =  0, 

that  of  P2P3  is  3x  —  2y  -I-  7  =  0, 

that  of  P3P4  is  X  +    y  +  4  =  0, 

and  tliat  of  P4P1  is  7x  —  2y  —  8  =  0. 

We  form  the  equation 

(x  -  3y  +  7)  (x  -1-  y+  4)  +  k  (3x  -  2y  +  7)  (7x  -  2y  -  8)  =  0, 
and,  substituting  the  coordinates  of  P5,  find 

k=:l- 

Hence,  the  required  conic  is 

(x  -  3y  +  7)  (X  +  y  +  4)  +  ^  (3x  -  2y  +  7)  (7x  -  2y  -  8)  =  0, 
or  109x2  -  I08xy  -|-  8y-  +  l(>i»x  -  lOy  -  108  =  0. 


270  PLANE    ANALYTIC    GEOMETRY. 

It  is  evident  that  the  above  method  is  always  applicable,  if 
no  three  points  lie  in  a  straight  line.     Hence  : 

Through  any  five  points,  no  three  of  which  lie  in  a  straight 
line,  one  and  only  one,  conic  may  he  drawn. 

If  three  of  the  points  lie  in  a  straight  line,  the  method  is 
applicable,  but  it  is  evident  that  the  conic  must  be  one  of  the 
limiting  cases,  for  it  must  consist  of  the  straight  line  in  which 
the  three  points  lie,  and  the  straight  line  connecting  the  other 
two  points. 

If  four  or  five  of  the  points  lie  in  a  straight  line,  the  method 
is  not  applicable.  It  is  geometrically  evident  that  in  this 
case  the  problem  is  indeterminate.  For  the  conic  may  consist 
of  the  straight  line  in  which  the  four  points  lie,  together 
with  any  line  through  the  fifth  point,  if  that  is  not  on  the  line 
with  the  four,  or  any  line  whatever  if  the  fifth  point  lies  on  a 
straight  line  with  the  four  others. 

If  it  is  required  to  determine  a  parabola,  only  four  points 
are  necessary.  This  follows  from  the  fact  that  one  rela- 
tion connecting  the  coefficients  is  always  given  ;  namely, 
AB  —  H^  =  0.  We  may  use  here,  also,  the  method  involving 
the  parameter  k,  if  no  three  of  the  points  are  in  the  same 
straight  line.  We  form,  as  before,  the  equation 
UiU3  +  kU2U,  =  0. 

We  form  then  the  equation  AB  —  H-  =  0  out  of  the  coeffi- 
cients of  this  equation.  The  result  is  a  quadratic  equation  in 
k,  and  hence  we  will  have  two,  one,  or  no  real  parabolas,  ac- 
cording as  the  values  of  k  are  real,  equal,  or  imaginary.  It 
should  be  noticed  that  in  this  connection  "parabola"  may 
mean  two  parallel  straight  lines. 

Ex.  Let  it  be  required  to  pass  a  parabola  through  the  points 
Pi  (1,-1),  P2(2,  3),  Pa  (2,  -5),  P4(5,  7).  We  find  the  equations  of 
the  following  lines  : 

P1P2,  4x  -  y  -  5  =  0  ;    P.Pg,  x  -  2  =  0  ;  P3P4,  4x  -  y  -  13  =  0; 
P4P1,  2x-y-3  =  0. 


THE   GENERAL   EQUATION.  271 

The  equation  of  the  conic  is,  then, 

{4x  -  y  -  5)  (4x  -  y  -  13)  +  k  (x  -  2)  (2x  -  y  -  3)  =  0, 
or 

(16  +  2k)  x-'^  +  (-  8  -  k)  xy  +  y^  +  (-  72  -  7k)  x  +  (18  +  2k)  y 
+  65  +  Gk  =  0. 
Since  this  is  to  be  a  parabola,  we  must  have 

(4  +  ^)'-(16  +  2k)  =  0, 

whence  follows  k  =  0  or  —  8. 

The  two  parabolas  are,  therefore, 

IGx-'  -  Bxy  +  y2  -72x  +  18y  +  G5  =  0, 
f  -  16x  +  2y  +  17  =  0. 
The  second  equation  represents  a  true  parabola  ;  the  first  factors  into 
(4x  -  y  -  5)  (4x  -  y  -  13)  =  0 
and  represents  two  parallel  straight  lines. 

EXAMPLES. 

Determine  the  nature  and  position  of  the  following  conies  : 

1.  4xy  +  3y2— 8x-16y  +  19  =  0. 

2.  x2  —  6xy  +  9y"  -  280x  —  20  ==  0. 

3.  11x2  —  4xy  +  Uf  —  26x  +  32y  +  59  =  0. 

4.  5x2  _  26xy  -f  5y2  +  lOx  -  26y  +  71  ==  0. 

5.  4xy  +  6x  — 8y  +  l=0. 

G.   x^  -  2xy  +  y2  +  2x  —  2y  +  1  =  0. 

7.  13x2  ^  loxy  +  13y2  +  6x  -  42y  —  27  =  0. 

8.  x2  -  4xy  -  2f  -  14x  -f  4y  +  25  =  0. 

9.  6x-'  —  5xy  —  fiy-'  —  46x  —  9y  +  GO  =  0. 

10.  4x2  —  8xy  +  4y2  +  Gx  -  8y  +  1  =  0. 

11.  x2  +  Gxy  +  9y2  -  Gx  -  18y  +  5  =  0. 

12.  41x2  —  24xy  +  34y2  —  188x  +  llGy  +  19G  =  0. 

13.  31x2  -  24xy  +  21y2  +  48x  -  84y  +  84  =  0. 

14.  Show  that  the  conic  represented  by  the  general  equa- 
tion is  an  equilateral  hyperbola  when  A  =  —  B. 

15.  Show  that,  if  the  general  equation  contains  the  term  in 
xy  and  not  more  than  one  of  the  terms  containing  x2  or  y-,  the 
conic  is  an  hyperbola. 


272  PLANE  ANALYTIC  GEOMETRY. 

16.  Prove  that  the  necessary  and  sufficient  conditions  that 
the  general  equation  shoukl  represent  a  circle  are  A  =  B, 
H  =  0,  provided  the  axes  are  rectangular. 

17.  Show  that  the  asymptotes  of  the  conic  are  parallel  to 
the  two  straight  lines 

Ax- 4- 2  Hxy+  By=^  =  0. 

18.  Show  that  if  A  and  B  have  opposite  signs  the  conic  is 
an  hyperbola. 

19.  Show  that  the  tangent  to  the  conic  at  the  point  (xj,  yi) 
is 

Axix  +  H(xiy  +  yix)  +  By^y  +  G(x  +  Xi)  +  F(y  +  y^j  +  C  =  0. 

Find  the  equations  of  the  conies  through  the  following 
points  : 

20.  (3,  2),  (-  2,  -  3),  a,  -  3),  (2.  -  2),  (|,  -  f ).      • 

21.  (1,-2),  (6,  3),  (3,2),  (2,1),  (9,  2). 

22.  (0,  a),  (a,  0),  (0,  -  a),  (-  a,  0),  (a,  a). 

23.  (1,1),  (-1,5),  (2,4),  (0,3),  (3,1). 

24.  Find  the  equation  of  a  parabola  through  the  four  points 
(5,-4),  (9,4),  (6,-1),  (51, -21). 

25.  Show  that  a  parallelogram  cannot  be  inscribed  in  a 
parabola. 

26.  Given  the  base,  2b,  of  a  triangle,  and  the  difference,  a, 
of  the  angles  at  the  base,  find  the  locus  of  the  vertex. 

27.  Given  the  base,  2b,  of  a  triangle,  and  the  difference,  d, 
of  the  tangents  of  the  angles  at  the  base,  find  the  locus  of  the 
vertex. 

28.  A  straight  line  has  its  extremities  on  the  coordinate  axes 
and  passes  through  a  fixed  point ;  find  the  locus  of  its  middle 
point. 


PART   II. 

SOLID  ANALYTIC  GEOMETRY. 


CHAPTER   I. 

THE    POINT. 

1.     Rectangular  Coordinates  in  Space. 

z 


N 

n' 

M 

X- 

/ 

F 

//' 

/ 

. — ^/. 

/  1 

/       1 

y                    1 
/ 
/ 
/ 

• 
• 

/ 

/ 

L                   ^ 

,/ 

M 

I 

Fig.  1. 


Let  XX',  YY',  and  ZZ'  be  three  lines  meeting  at  0  so  that 
each  is  perpendicular  to  the  other  two.     They  will  determine 


274  SOLID    ANALYTIC    GEOMETRY. 

three  plaues,  each  of  which  will  be  perpendicular  to  the  other 
two,  and  which  will  be  called  the  coordinate  planes,  the  three 
lines  XX',  YY',  and  ZZ'  being  called  the  axes  of  x,  y,  and  z, 
respectively,  or  the  coordinate  axes. 

Distances  from  the  plane  YOZ  measured  parallel  to  XX' 
shall  be  denoted  by  x,  x  being  positive  if  measured  in  the 
direction  OX,  and  negative  if  measured  in  the  direction  OX'. 
Similar  meanings  are  assigned  to  y  and  z.  There  can  then 
be  but  one  set  of  simultaneous  values  of  x,  y,  and  z  corre- 
sponding to  any  given  point.  For  through  the  point  only 
one  plane  can  be  passed  parallel  to  the  plane  YOZ,  and  the 
distance  of  this  plane  from  the  plane  YOZ,  measured  accord- 
ing to  the  rule  above,  will  be  the  value  of  x  corresponding  to 
the  point.  Similarly,  there  will  be  but  one  value  of  y  and  one 
value  of  z  corresponding  to  the  point. 

On  the  other  hand,  a  point  is  completely  determined  if 
simultaneous  values  of  x,  y,  and  z  for  the  point  are  known. 
For  example,  let  x  =  a,  y  =  b,  z  —  c.  If  x  =  a,  the  point  is 
in  a  plane  parallel  to  the  plane  YOZ,  and  a  units  distant 
from  it.  If  y  =  b,  the  point  is  in  a  plane  parallel  to  the 
plane  XOZ,  and  b  units  from  it.  If  z  ==  c,  the  point  is  in 
a  plane  parallel  to  the  plane  XOY,  and  c  units  from  it. 
These  three  planes  meet  in  one  and  only  one  point,  which 
has  for  coordinates  x  =  a,  y  =  b,  z  =  c,  or  the  point  (a,  b,  c). 

Therefore  we  see  that,  given  the  point,  there  can  be  but 
one  set  of  coordinates  corresponding  to  it,  and  that,  given  the 
coordinates,  there  can  be  but  one  point  corresponding  to  them. 

In  Fig.  1,  OL,  ML',  M'P,  NN'  all  represent  x;  OM,  NM', 
N'P,  LL'  all  represent  y;  and  ON,  MM',  L'P,  LN'  all  repre- 
sent z. 

It  follows  that  the  point  may  be  plotted  in  several  ways, 
ars  represented  in  Fig.  2  ;  but  the  first  method  shown  is  the 
one  most  often  chosen. 

If  the  axes  are  not  mutually  perpendicular  to  each  other, 
the  coordinates  are  called  oblique,  but  the  rules  for  their 


THE    POINT. 


275 


Fig.  2. 


measurement  and   the  plotting  of   points  are  the   same  as 
those  just  given  for  rectangular  coordinates. 


2.     Polar  Coordinates  in  Space. 

In  §  1  we  have  determined  tlie  position  of  a  i)oint  by- 
means  of  its  three  distances  from  the  three  coordinate  planes 
respectively.  In  this  article  we  are  to  determine  the  posi- 
tion of  the  point  by  means  of  its  distance  from  a  fixed  point 
and  the  direction  in  which  this  distance  is  measured. 

Let  0  (Fig.  3)  be  the  fixed  point,  or  origin,  and  OX  and  OZ 
be  two  lines  of  reference  at  right  angles  to  each  other  and 
determining  a  plane  XOZ.  The  distance  OP  of  the  point 
from  the  origin  shall  be  called  r.     The  direction  of  OP  will 


276 


SOLID    ANALYTIC    GEOMETRY. 


Fig.  3. 


be  determined  if  tlie  plane  in  which  it  lies  and  the  angle  it 
makes  with  OZ  are  known.  The  angle  ZOP  shall  be  called 
(f),  and  the  diedral  angle  between  the  planes  ZOX  and  ZOP 
shall  be  called  6.  We  now  have  the  point  completely  deter- 
mined, its  coordinates  being  (r,  <f>,  6). 

At  first  it  may  not  be  evident  that  in  polar  coordinates  the  point  is 
determined  by  the  intersection  of  three  surfaces,  yet  such  is  the  case. 
For,  if  0  is  given,  the  point  is  in  one  of  the  planes  radiating  from  OZ. 
If  <j)  is  given,  the  point  is  on  the  surface  of  a  cone  of  revolution  of  which 
the  axis  is  OZ,  and  the  elements  of  which  make  the  angle  </>  with  the  axis. 
These  two  surfaces  intersect  in  a  straight  line  radiating  from  0.  Finally, 
if  r  is  given,  the  point  is  on  the  surface  of  a  sphere  having  its  centre  at  0, 
and  is  the  point  where  the  spherical  surface  is  pierced  by  the  above  line 
of  intersection.      (See  Fig.  4.) 


<^  will  vary  from  0°  to  180°,  0  from  0°  to  360°,  and  r  can 
have  all  numerical  values.  If  we  wish  to  give  r  a  negative 
value,  we  may  do  so  as  follows  :  As  noted  above,  the  coordi- 
nates <^  and  $  determine  a  straight  line  radiating  from  0,  and 
r  will  be  positive  if  the  point  is  on  this  line,  and  negative 
if  the  point  is  on  the  extension  of  this  line  backward  through 
the  origin. 


277 


^ 


Fig.  4. 


3.     Distance  between  Two  Points  Pi  (xi,  yi,  Zj)  and  Pj  (xj, 
I2,  22)- 

Z 


Draw  Pi  Ml  and  P2M2  parallel  to  OZ  and  meeting  the  plane 
XOY  at  Ml  and  M2,  respectively,  and  MiLj  and  MgLg  parallel 


278 


SOLID    ANALYTIC    GEOMETRY. 


to  OY  and  meeting  OX  at  Lj  and  Lj,  respectively.  Then 
OLi  =  xi,  LiMi  =  yi,  MiPi  =  Zj,  etc.  In  the  plane  XOY  draw- 
Mo  R  parallel  to  OX,  and  in  the  plane  determined  by  the  par- 
allel lines  PiMi  and  P2M2  draw  P2S  parallel  to  M1M2,  the 
line  of  intersection  of  this  plane  with  the  plane  XOY.  Then 
P1P2  and  M1M2  are  the  respective  hypotenuses  of  the  right 
triangles  PiSPg  and  MiRMo. 


Therefore  Pi  P2'  =  S  P2'  +  S  Pi 

=  ivuvi;'+sp;' 


=  RM2"+  RMr+SPi 
-  (xi  -  X2)^  +  (yi  -  y,f  +  (zi  -  Z2)^ 
for  RM2  =  X2  —  Xi,  RMi  =  yi  —  ya,  and  SPi  =  Zi  —  z^. 
Or,  if  PiP2  =  d, 

d  =  ^y(Xl  -  x,y  +  (yi  -  y^f  +  (Zi  -  Z2/. 


[1] 


4.  To  find  the  Coordinates  of  a  Point  P  (x,  y,  z)  which 
divides  the  Line  joining  Pi  (xi,  yi,  Zi)  and  P2  (Xo,  Y2,  Z2)  so 
that  PiP  :  PPo  =  li :  L. 

z 


Fig.  6. 


THE    POINT.  279 

To  find  X,  draw  PiNi,  PN,  and  P2N2  parallel  to  OX  and 
meeting  the  plane  YOZ  at  Ni,  N,  and  Ng,  respectively.  In  the 
plane  determined  by  the  parallel  lines  PiNi  and  PgNo  draw 
PiR  and  PS  parallel  to  NiNN.,  the  line  of  intersection  of 
this  plane  with  the  plane  YOZ.  Then  NiPi  =  Xj,  NP  =  x, 
and  N2P2  =  Xo,  so  that  RP  =  x  —  x^  and  SPo  =  X2  —  x. 

The  triangles  PiRP  and  PSPo  are  similar,  since  their  sides 
are  respectively  parallel. 

PiP_  RP 
PP.,~SP.,' 


Therefore 


or 


ll  X  —  Xi 


whence  X  = 


Xo  —  X 

1 1X2  +  I2X1 

ll  +  I. 


Similarly,     y  =      \   ^\      '  ^^^^^  ^  ^  "[^^^TJ^  *  ^2] 


5.     Direction  Cosines. 

The  angle  between  any  two  lines  is  the  angle  between  their 
respective  parallels  drawn  from  a  common  point.  Hence  any 
line  in  space  may  be  regarded  as  making  angles  a,  ft,  and  y, 
respectively,  with  the  axes  of  x,  y,  and  z,  these  being  the 
angles  made  with  the  axes  by  the  line  through  the  origin 
parallel  to  the  given  line.  Then  cos  a,  cos  j8,  cos  y  are  called 
the  direction  cosines  of  the  line. 

If  P  (x,  y,  z).  Fig.  7,  is  any  point,  and  0P  =  r,  and  the 
direction  cosines  of  OP  are  cos  a,  cos  y8,  cos  y,  by  Trigonom- 
etry OL  ==  OP  cos  a,  OM  =:  OP  cos  /?,  and  ON  =  OP  cos  y,  or 

x  =  r  cos  a,  y  =  r  cos  13,  z  ^  r  cos  y. 

Squaring  and  adding,  we  have 

X"  +  y'  +  z"  =  r^  (cos^  a  -\-  cos'-  /5  +  cos'-'  y). 


280 


SOLID   ANALYTIC    GEOMETRY. 


oy. 


Fig.  7. 

But  by  [1],  x2  +  y-  +  z2=r^; 

.•.  cos^  a  +  cos'-^  P  +  cos-  7  =  1,  [3] 

a  fundamental  relation  between  the  three  direction  cosines  of 
any  line. 

The  quantities,  r,  cos  a,  cos  (3,  cos  y,  may  be  regarded  as 
the  coordinates  of  a  point,  since  they  are  sufficient  to  fix  it 
uniquely. 

Any  three  quantities,  I,  m,  and  n,  are  proportional  to  the 
direction  cosines  of  some  line ;  for  if 

cos  a cos  (3 cos  y 

I  m  n     ' 

and  the  value  of  this  constant  ratio  be  denoted  by  r,  we  have 

cos  a  =  Ir,  cos  /?  ^  mr,  cos  y  :=  nr, 

whence,  by  substituting  in  [3],  we  find 


V>  +  m^  +  n= 


THE    POINT. 


281 


SO  that 


cos  a 


cos  /3  =  ^  =^, 

Vl^+  m"  -|-  n- 


cos  y  ^ 


VP  +  m^  +  n'^ 


6.     Projection. 

The  projection  of  one  line  on  a  second  line  is  the  portion 
of  the  second  line  included  between  two  planes  passed  through 
the  respective  ends  of  the  first  line  perpendicular  to  the  second. 

1.  The  projection  of  one  line  on  a  second  is  the  product  of  the 
first  line  hy  the  cosine  of  the  angle  between  the  lines. 

R 
M 


Fig.  8. 

Let  AB  be  any  line,  and  MN  and  RS  be  planes  through  A 
and  B,  respectively,  perpendicular  to  the  line  C  D  and  meeting 
CD  at  the  points  A'  and  B',  respectively.  Then  A'B'  is  the 
projection  of  AB  on  CD,  and  if  6  denote  the  angle  between 
AB  and  CD,  we  wish  to  prove  A'B'  =  AB  cos  $. 

Erom  A'  draw  A'E  parallel  to  AB  and  meeting  the  plane 
RS  at  E.  Then  A'E  erpials  AB,  since  they  are  parallel  lines 
included  between  parallel  planes.  Biit,  by  Trigonometry, 
A'B' =  A'E  cos  ^ ;  therefore  A'B' ^  AB  cos  ^,  as  we  wished 
to  prove. 


282 


SOLID    ANALYTIC    GEOMETRY. 


If  we  define  the  projection  of  a  broken  line  on  a  given 
straight  line  as  the  algebraic  sum  of  the  projections  of  its 
segments  on  that  line,  we  shall  have,  as  a  second  theorem  : 

2.  The  projections  on  any  given  line  of  a  broken  line  and  of 
a  straight  line  joini7ig  the  ends  of  the  broken  line  are  the  same. 


Fig.  9. 


For  example,  the  projection  of  the  broken  line  ABCDE  on 
MN  is  A'B'+  B'C'  +  C'D'+  D'E',  which  equals  A'E',  and  A'E'  is 
the  projection  of  the  straight  line  AE  on  MN. 


7.     Angle  between  Two  Lines. 

Let  OM  and  ON  (Fig.  10)  be  two  lines  having  as  direction 
cosines  cos  ai,  cos  ^i,  cos  yi,  and  cos  ag,  cos  /?2,  cos  yo,  re- 
spectively. Let  angle  MON  be  denoted  by  6.  Take  P  (xi,  yi,  z^) 
any  point  of  OM,  distant  ri  from  0. 

By  1,  §  6,  the  projection  of  OP  on  ON  is  ri  cos  6,  and  the 
projection  of  the  broken  line  OLMP  on  ON  is 

OL  cos  a.,  +  LM  cos  /3.J  +  M  P  cos  ya, 

or  Xi  cos  aa  +  Yi  cos  ^2,  +  Zi  cos  yo. 

Therefore,  by  2,  §  6, 

Xi  cos  6=Xi  cos  a2  +  Yi  cos  fi^  +  z^  cos  y^, 

whence        cos  6^~  cos  a,  +  ^  cos  /3o  +  —  cos  y^, 
1-1  '       ri         ^        ri         ' 


THE    POINT. 


283 


Fig.  10. 


or       cos  9  =  cos  ai  cos  ao  +  cos  Pi  cos  Pa  +  cos  7i  cos  y^,         [4] 
since  by  §  5,  —  =  cos  aj,  etc. 


As  the  angle  between  any  two  lines  is  the  angle  between 
their  respective  parallels  drawn  from  a  common  point,  [4]  is 
obviously  true  in  all  cases. 

If  ^  =  90°,  cos  ^  =;  0.  Therefore,  if  two  lines  are  perpen- 
dicular to  each  other, 

cos  ai  cos  tto  4-  cos  Pi  cos  po  -\-  cos  "Yi  cos  y.,  =  0.  [5] 

If  two  lines  are  parallel,  it  is  evident  that 

ai  =  tto,       pi  =  po,       \i  =  V2-  [6] 


8.  Transformation  of  Coordinates,  the  New  Axes  being 
respectively  Parallel  to  the  Old  Axes. 

Let  OX,  OY,  OZ  denote  the  original  axes,  and  0'X\  O'Y', 
O'Z'  denote  the  new  axes,  the  coordinates  of  0'  with  respect 
to  the  original  axes  being  Xq,  yo,  Zo- 


284 


SOLID    ANALYTIC    GEOMETEY. 


Let  P  be  any  point  in  space,  its  original  coordinates  being 
X,  y,  z,  its  new  being  x',  y',  z'.  From  P  draw  a  line  parallel 
to  OX,  and  hence  parallel  to  O'X',  and  meeting  the  planes 
YOZ  and  Y'O'Z'  at  the  points  M  and  M',  respectively.  Then, 
by  definition,  MP  =  x  and  M'P  =  x';  also,  since  parallel 
lines  included  between  parallel  planes   are  equal,  MM'  =  Xo. 

But  by  (1),  §  1,  Part  I, 

MP=MM'+  M'P. 


Similarly, 


.-.      X  =  Xo  +  X'. 

y  =  Yo  +  y'»     z  =  Zo  +  z'. 


[7] 


9.  Transformation  of  Coordinates  from  One  Set  of  Rect- 
angular Axes  to  a  New  Set  of  Rectangular  Axes  having  the 
Same  Origin. 


.^'X' 


.,-Hr7 


'M' 


/L 


Fig.  11. 


In  iFig.  11,  OX,  OY,  and  OZ  are  the  original  axes,  and  OX', 
OY',  and  OZ'  are  the  new  axes,  their  direction  cosines  with 


THE    POINT.  285 

respect  to  OX,  OY,  and  OZ  being  respectively  cos  ai,  cos  fii, 

cos  yi,  cos  a^,   COS  jSg,  COS  y^,   and  cos  as,  cos  jSs,  cos  ys. 

Let  P  be  any  point  in  space,  its  old  coordinates  being 
X,  y,  z,  its  new  being  x',  y',  z'.  From  P  draw  PM  parallel  to 
OZ  and  meeting  the  plane  XOY  at  M,  and  draw  ML  parallel 
to  OY  and  meeting  OX  at  L.  Also  draw  PM'  parallel  to  OZ' 
and  meeting  the  plane  X'OY'  at  M',  and  draw  M'L'  parallel  to 
OY'  and  meeting  OX'  at  L'. 

Then  OL  =  x,  LM  ==  y,  M P  ==  z, 

OL'  =  x',  L'M'  =  y',  M'P  =  z'. 

Now  the  projection  of  OP  on  OX  is  OL  or  x,  and  by  2,  §  6, 
this  is  the  projection  of  the  broken  line  OL'M'P  on  OX. 

The  projection  of  OL'  on  OX  is  OL'cos  ai,  for  the  angle 
between  OL'  and  OX  is  aj.  The  projection  on  OX  of  L'M'  is 
L'M'  cos  aa,  for  L'M'  is  parallel  to  OY'  and  the  angle  between 
OY'  and  OX  is  aj.  Similarly,  the  projection  on  OX  of  M'P  is 
M'P  cos  as.  Therefore,  by  §  6,  the  projection  on  OX  of 
OL'M'P  is 

OL'  cos  ai  +  L'M'  cos  a.,  +  M'P  cos  as, 
so  that  finally 

X  =  x'  cos  tti  +  y'  cos  a.  +  z'  cos  a-^. 
Similarly, 

y  =  x'  cos  Pi  +  y'  cos  Po  +  z'  cos  ps,  [8] 

z  =  x'  cos  "Yi  +  y'  cos  y.,  +  z'  cos  ^3. 

In  similar  manner,  we  can  deduce  the  following  expressions 
for  x',  y',  z'  in  tei'ms  of  x,  y,  and  z  : 

x'  ^  X  cos  ai  +  y  cos  ^1  +  z  cos  yi, 
y'=  X  cos  as  +  y  cos  (So  -\-  z  cos  yo, 
z'  :=  X  cos  a;5  +  y  cos  jSs  +  z  cos  ys. 

As  was  to  be  expected,  these  formulas  are  of  the  same 
form  as  [8],  for  the  two  transformations  are  of  the  same 
kind. 


286  SOLID    ANALYTIC    GEOMETRY. 

In  using  these  formulas  the  student  should  bear  in  mind 
that  the  axes  are  perpendicular  to  each  other,  and  hence  that 

cos  tti  cos  a2  +  cos  (3i  cos  ^2  +  cos  yi  cos  y2  =  O5  ^ 
cos  aa  COS  as  -|-  COS  (S^  COS  /Sg  +  COS  y^  COS  yg  =  0,  I 
COS  ag  COS  tti  +  COS  jSg  COS  ySi  +  COS  yg  COS  y^  =  0 ;  J 

also  that 

cos  ai  COS  )8i  +  COS  ag  COS  /Jg  +  COS  ag  COS  /Sg  :=  0,  '^ 
COS  /?i  COS  yi  +  COS  ^2  COS  ya  +  COS  jSg  COS  ys  =  0,  I 
COS  yi  COS  tti  +  COS  y2  COS  a2  +  COS  yg  COS  ag  =  0  ;  I 

also,  by  [3],  that 

cos-  ai  +  COS^  I3i  +  COS^  yi  =  1, 

cos^  ttg  +  cos^  ^2  +  cos^  y2  =  1; 

COS^  ag  +  COS^  ygg  +  COS^  yg  =  1 ;  j 

and  COS^  ai  +  COS^  ao  +  COS-  ag  =  1,^ 

cos^  /3i  +  COS- 132  +  cos^  /Sg  =  1,  J> 

COS^  yi  +  COS^  y2  +  COS^  yg  =  1.  J 

10.  All  transformations  involving  only  rectangular  co- 
ordinates can  be  made  by  the  use  of  one  or  both  of  these 
sets  of  formulas,  as  the  case  may  be ;  and  since  both  sets  give 
the  original  coordinates  as  expressions  of  the  first  degree  in 
terras  of  the  new,  it  follows  that  the  degree  of  an  equation  is 
not  changed  by  any  such  transformation.  The  proof  is  like 
that  given  in  §  48,  Part  I,  for  the  coordinates  of  the  point  in 
the  plane,  and  for  tliat  reason  need  not  be  given  here,  though 
the  student  is  advised  to  write  out  the  proof  for  himself. 

11.  Relation  between  Rectangular  and  Polar  Coordinates. 

If  the  lines  OX  and  OZ  used  in  defining  the  polar  coordi- 
nates of  a  point  are  respectively  the  axes  OX  and  OZ  of 
rectangular  coordinates,  it  is  not  difficult  to  express  the 
rectangular  coordinates  of  any  point  in  terms  of  its  polar 
coordinates,  and  vice  versa. 


THE    POINT. 


287 


Fig.  12. 

For  if,  in  Fig.  12,  the  polar  coordinates  of  P  (x,  y,  z)  are 
r,  (j>,  and  0,  and  PM  is  drawn  parallel  to  OZ,  it  is  readily  seen 
that  OP  =  r,  Z  ZOP  =  </>,  and  Z  XOM  =  0,  since  Z  XOM  is 
the  plane  angle  of  the  diedral  angle  between  the  planes  XOZ 
and  MONP. 


Therefore 
and 
also 
and 


ON  =  OP  cos  <^, 
OM  =  OP  sin  <^; 
LM  =  OM  sin  0, 
OL  =  OM  cos  0. 


By  substitution  of  the  respective  values  of  OP,  ON,  OM,  L^ 
and  OL  in  these  equations  we  get,  as  our  desired  equations. 


z  =  r  cos  (|), 

X  =  r  sin  <j)  cos  9, 

y  =  r  sin  (j)  sin  0. 


[9] 


From  these  formulas  may  be  deduced  the  following,  which 
express  the  polar  coordinates  of  any  point  in  terms  of  its 
rectangular  coordinates  : 


288  SOLID    ANALYTIC    GEOMETRY. 


r  —  Vx"  +  y^  +  z^, 

EXAMPLES. 

1.  What  is  the  distance  from  (—  3,  4,  5)  to  (3,  —  4,-2)? 

2.  Find  the  points  of  trisection  of  the  line  joining 
(2,  3,  —  4)  to  (5,  6,  -  10). 

3.  The  line  joining  (xj,  y^,  Zi)  to  (x^,  y^,  Zg)  is  divided  exter- 
nally in  the  ratio  l^  :  Ig.  Find  the  coordinates  of  the  point  of 
division. 

4.  Find  the  coordinates  of  the  point  dividing  the  line 
passing  through  (1,  4,  —  3)  and  (2,  —  3,  5)  in  the  ratio  3  :  —  2. 

5.  The  line  joining  the  point  (1,  4,  1)  to  (3,  —  1,  3)  is 
extended  in  the  same  direction  until  its  length  is  doubled. 
What  are  the  coordinates  of  the  point  reached  ? 

6.  Find  the  coordinates  of  the  points  at  a  distance  5  from 
the  points  (—  2,  —  2,  1),  (3,  —  2,  6),  and  (3,  3,  1). 

7.  Find  the  radius  and  the  centre  of  the  sphere  having  the 
line  joining  ( — 4,  1,  7)  and  (2,  — 3,  9)  as  a  diameter. 

8.  Find  the  radius  and  the  centre  of  the  sphere  passing 
through  the  points  (0,  0,  0),  (1,-2,-2),  (2,  1,  3),  and 
(3,  1,  -  1). 

9.  A  line  makes  equal  angles  with  the  three  axes.  What 
is  that  angle  ? 

10.  Find  the  direction  cosines  of  a  line  which  makes  an 
angle  of  30°  with  the  axis  of  x  and  equal  angles  with  the 
axes  of  y  and  z. 

11.  The  direction  cosines  of  a  line  are  proportional  to  3, 
—  4,  and  2.     Find  their  values. 

12.  Find  the  angle  between  the  lines  joining  the  origin  to 
the  points  (2,  —  3,  1)  and  (—  2,  3,  —  1). 


THE    POINT.  289 

13.  Find  the  rectangular  coordinates  of  the  point  (3,  60°, 
30°). 

14.  Find  the  polar  coordinates  of  the  point  (3,  12,  4). 

15.  Show  that  the  distance  between  the  points  (rj,  ^i,  Oi) 
and  (ra,  cf)2,  62)  is 

Vri^  +  r^^  —  2rir2  [sin  <^i  sin  <^2  cos  (^1  —  $2)  +  cos  <^i  cos  cjba]. 

16.  If  0  is  the  angle  between  two  lines,  prove 

sin"  0  ==  (cos  tti  cos  ^2  ~  cos  ag  cos  /?i)^ 

+  (cos  {3i  cos  yo  —  cos  /Sg  cos  yi)^ 
+  (cos  yx  COS  a2  —  COS  y2  COS  ai)^. 

17.  In  any  quadrilateral  the  sum  of  the  squares  of  the 
diagonals  is  double  the  sum  of  the  squares  of  the  lines  join- 
ing the  middle  points  of  the  opposite  sides.     Prove. 

18.  Show  analytically  that  the  lines  joining  the  middle 
points  of  the  opposite  sides  of  any  quadrilateral  are  bisected 
at  a  common  point. 

19.  Prove  analytically  that  the  straight  lines  joining  the 
middle  points  of  the  opposite  edges  of  a  tetraedon  are 
bisected  at  a  common  point. 

20.  Prove  analytically  that  the  lines  joining  each  vertex  of 
a  tetraedon  with  the  point  of  intersection  of  the  medial  lines 
of  the  opposite  face  all  meet  in  a  point  which  divides  each 
line  in  the  ratio  of  3  : 1. 

21.  If  E,  F,  G,  H  are  the  middle  points  of  tlie  edges  AB, 
AD,  CD,  and  BC,  respectively,  of  the  tetraedon  A  BCD,  prove 
that  EFGH  is  a  parallelogram. 


CHAPTER   II. 

INTERPRETATION    OF    EQUATIONS. 

12.     Single  Equation. 

Let  lis  consider  certain  examples  of  a  single  equation  involv- 
ing the  coordinates  x,  y,  z. 

1.  The  simplest  single  equation  is  of  the  form  x  =  a.  This 
equation  is  evidently  satisfied  by  the  coordinates  of  every 
point  in  a  plane  parallel  to  the  plane  YOZ  and  a  units  distant 
from  it,  and  by  the  coordinates  of  no  other  point.  Therefore, 
the  equation  x  =  a  represents  a  plane. 

2.  Consider  secondly  the  equation 

Ax+  By  +  Cz+  D  =  0. 

If  we  assume  values  for  x  and  z,  respectively,  as  xi  and  Zj, 
and  substitute  in  the  equation,  we  shall  have  the  equation 

Axi+  By  +  Czi+D  =  0, 

from  which  we  can  determine  y.  In  this  way  we  shall  find  a 
point  of  the  locus  of  this  equation  of  which  the  projection 
upon  the  plane  XOZ  is  the  point  (xi,  Zi).  But  we  may  give  x 
and  z  any  values  we  choose  and  in  every  case  we  shall  find  one 
and  only  one  corresponding  point  of  the  locus.  Hence  the 
locus  has  extension  in  two  dimensions,  but  no  thickness,  and 
is  therefore  a  surface  of  some  kind. 

We  shall  prove  in  the  following  chapter  that  the  surface  is 
a  plane. 

3.  As  a  third  example,  we  will  take  the  equation 

x^  +  y2  +  z^  =  al 


INTERPRETATION  OF  EQUATIONS.         291 

Placing  X  =^  Xi,  and  z  =  zi,  we  find  for  the  corresponding 
values  of  y, 

y  =  ±Va'— (xi^  +  zi^). 

If,  now,  we  take  Xj,  Zi,  so  tliat  xf'  +  z^^  >  a-,  we  can  find  no 
corresponding  point  on  the  locus,  for  y  becomes  imaginary. 
But  if  Xi"-^  +  zi^  <  a^,  i.e.,  if  (xj,  Zj)  is  within  the  circumference 
of  the  circle  with  radius  a  which  has  its  centre  at  the  origin, 
we  find  two  corresponding  points  of  the  locus.  If  Xi^  +  z^^  =  a^, 
i.e.,  if  (xi,  Zj)  is  on  the  circumference  of  the  above-named 
circle,  the  two  values  of  y  coincide.  Since,  then,  (xi,  Zi)  may 
be  taken  anywhere  within  or  on  the  circle,  it  follows  that  the 
locus  has  extension  in  two  dimensions.  It  has,  however,  no 
thickness,  for  the  two  values  of  y  corresponding  to  (xi,  z^)  are 
isolated,  and  the  points  between  them  do  not  belong  to  the 
locus.  Hence  the  locus  is  a  surface.  A  reference  to  §  3 
shows  that  this  surface  is  in  fact  a  sphere  with  radius  a  and 
centre  at  the  origin. 

4.  It  is  evident  that  the  above  reasoning  may  be  applied  to 
any  given  equation.  For  we  may  give  to  x  and  z  any  two 
values  (xi,  Zj),  thus  obtaining  for  the  locus  extension  in  two 
dimensions.  We  compute,  then,  from  the  equation  the  cor- 
responding values  of  y,  which  may  be  n  in  number.  There 
are,  then,  n  points  of  the  locus  perpendicularly  above  or 
below  the  xz-plane,  but  these  points  are  isolated  and  the 
space  between  them  does  not  belong  to  the  locus.  Hence  the 
locus  has  no  thickness.     Therefore,  we  may  say  : 

A  single  equation  involving  the  coonllnates  of  a  j^oint  repre- 
sents a  surface. 

There  are  apparent  exceptions  to  the  above  tlieorem,  if  we  demand 
that  the  surface  shall  have  real  existence.     Thus,  for  example, 

x2  -f-  y2  -|-  22  z=  —  1 

is  satisfied  by  no  real  values  of  the  coordinates.  It  is  convenient  in  sucli 
cases,  however,  to  speak  of  ' '  imaginary  surfaces. ' ' 


292  SOLID    ANALYTIC    GEOMETRY. 

Moreover,  it  may  happen  that  the  real  coordinates  which  satisfy  the 
equation  may  give  points  which  lie  upon  a  certain  line,  or  are  even 
isolated  points.     For  example,  the  equation 

x2  +  y2  =  0 
is  satisfied  in  real  coordinates  only  by  the  points  (0,  0,  z),  which  lie  upon 
the  axis  of  z ;  while  the  equation 

X2  +  y2  +  z-^  =  0 

is  satisfied,  as  far  as  real  points  go,  only  by  (0,  0,  0).  In  such  cases,  it  is 
still  convenient  to  speak  of  a  surface  as  represented  by  the  equation,  and 
to  consider  the  part  which  may  be  actually  constructed  as  the  real  part 
of  that  surface.  The  imaginary  part  is  considered  as  made  up  of  the 
points  corresponding  to  sets  of  imaginary  expressions  in  x,  y,  and  z, 
which  satisfy  the  equation.  With  this  understanding  the  general  dis- 
cussion (4)  may  be  read  without  change,  there  being  no  distinction  made 
between  real  and  imaginary  quantities. 

13.     Two  or  More  Equations. 

Two  simultaneous  equations  represent  a  line. 

When  two  simultaneous  equations  are  given,  it  is  under- 
stood tliat  the  locus  consists  of  those  points,  the  coordinates 
of  which  satisfy  both  equations.  Now,  any  point,  the  coordi- 
nates of  which  satisfy  either  equation,  must  be  on  the  surface 
represented  by  that  equation.  Hence,  any  point,  the  coordi- 
nates of  which  satisfy  both  equations  at  the  same  time,  must 
be  in  each  of  the  surfaces  represented  by  those  equations,  i.e., 
in  their  line  of  intersection.  Therefore,  two  equations  repre- 
sent the  line  of  intersection  of  the  two  surfaces  respectively 
represented  by  the  equations  when  taken  separately. 

As  in  the  previous  article,  it  may  happen  that  it  is  necessary  to  speak 
of  "imaginary  lines."  This  happens  when  the  real  parts  of  the  two  sur- 
faces represented  by  the  two  equations  do  not  intersect.  In  exceptional 
cases,  also,  the  line  of  intersection  contains  only  isolated  real  points. 
Such  would  be  the  case  if  either  surface  were  of  the  kind  mentioned  in 
the  latter  part  of  the  fine  type  of  the  previous  article. 

Three  independent  equations  taken  simultaneously  determine 
a  mimber  of  points. 


INTERPRETATION    OF    EQUATIONS.  293 

For  the  sets  of  values  of  x,  y,  and  z,  which  satisfy  three 
simultaneous  equations,  are  known,  by  Algebra,  to  be  limited 
in  number,  provided  the  three  equations  are  independent. 
Geometrically,  these  values  correspond  to  the  points  which 
the  three  surfaces  represented  by  the  equations  have  in  com- 
mon. These  points  may  be  found  by  noticing  the  points  in 
which  the  line  of  intersection  of  the  first  two  surfaces  meets 
the  third  surface.  It  is  not  necessary  that  all  or  any  of  these 
points  should  be  real. 

14.     Cylinders. 

If  a  given  equation  involves  only  two  of  the  coordinates, 
it  might  appear  to  the  student  to  represent  a  curve  lying  in 
the  plane  of  those  two  coordinates.  For  example,  the  equa- 
tion x^  +  y^  =  a"'^  might  appear  to  represent  an  equation  in  the 
plane  XOY.  Such  an  interpretation  of  the  equation  would  be 
incorrect,  for  it  amounts  to  restricting  z  to  the  value  z  =  0 ; 
whereas,  in  fact,  the  value  of  z,  corresponding  to  two  values 
of  X  and  y  which  satisfy  the  equation,  may  be  anything  what- 
ever. Hence,  corresponding  to  a  point  of  the  circle  in  the 
plane  XOY,  there  is  an  entire  straight  line"  in  space,  parallel 
to  OZ  and  lying  on  the  surface  represented  by  x-  -|-  y-  =  a^. 
Hence  this  surface  is  a  right  circular  cylinder. 

In  like  manner,  z^  =  4py  is  a  parabolic  cylinder,  the  ele- 
ments of  which  are  parallel  to  OX.     So,  in  general  : 

An  equation  containing  only  two  coordinates  represents  a 
cylindrical  surface,  the  elements  of  ivhich  are  jparallel  to  the 
axis  of  the  missing  coordinate. 

If  only  one  coordinate  is  present  in  the  equation,  as,  for 
example,  in  x'''  —  (a  +  b)  x  -f  ab  =  0,  the  locus  is  a  number  of 
planes. 

The  above  equation  may  be  written 

(x  —  a)  (x  —  b)  =  0, 


294  PLANE   ANALYTIC   GEOMETRY. 

and  hence  by  reasoning  like  that  of  §  40,  Part  I,  it  represents 
the  planes 

X  —  a  =  0  and  x  —  b  ^  0. 

It  is  evident  that  any  equation  involving  only  one  coordinate 
can  always  be  treated  in  this  way. 

As  a  plane  is  a  cylindrical  surface  of  which  the  directrix  is 
a  straight  line,  we  may  say  in  general  that  any  equation  not 
contaiyiing  all  three  of  the  coljrdmates  rejivesents  a  cylindrical 
surface. 

If  the  axes  are  oblique,  this  class  of  equations  will  repre- 
sent cylindrical  surfaces  of  which  the  elements  are  not  per- 
pendicular to  the  plane  of  the  directrix. 


15.     Surfaces  of  Revolution. 

A  surface  formed  by  the  revolution  of  a  plane  curve  about 
one  of  the  coordinate  axes  is  called  a  surface  of  revolution. 
The  equation  of-  such  a  surface  is  readily  formed  from  the 
equation  of  the  plane  curve  which  is  rotated  to  form  it.  By 
a  few  examples  we  will  show  how  to  do  this,  and  at  the  same 
time  we  will  deduce  a  rule  by  which  we  can  often  tell  by  in- 
spection that  a  given  equation  represents  a  surface  of  revo- 
lution. 

1.  The  circle  x^  +  z^=a^  Fig.  13,  is  rotated  about  OX  as 
an  axis,  forming  the  sphere  of  radius  a. 

If  Pi  is  a  point  of  the  circle,  then  the  equation  of  the  circle 
may  be  written 

OL'  +  LPT  =  a-". 

As  the  circle  is  rotated  about  OX,  OL  remains  fixed,  but 
LPi  takes  the  position  LP.     Hence  we  have  for  the  point  P, 

Ol'  +  LP'  =  a^. 


INTERPEETATION    OF    EQUATIONS. 


295 


Fig.  13. 


If,  now,   X,   y,   z  are  the  coordinates   of    P,   OL  =  x    and 
LP  =  Vy^+  z^.     Therefore  the  equation  of  the  sphere  is 

x^  +  y^  +  z"  =  a^- 

Here  we  have  formed  the  equation  of  the  sphere  from  tliat 
of  the  circle  by  replacing  z  by  Vy^  +  z^. 


2.  The  hyperbola  —  —  -"  =  1,  Fig.  14,  is  rotated  about  OX. 
The  equation  of  the  hyperbola  is 

OL'  _  Lp;'  ^ 


296 


SOLID   ANALYTIC    GEOMETRY. 

Z 


Fig.  14. 

When   Pi  is  rotated  into  the  position   P,  OL  is  fixed  and 
LPi  becomes  LP. 

OL'       LP'  _  -. 


But  OL  =  x,  and  LP=:  Vy"  +  z^.     Therefore  the  required 

equation  is 

x;  _  y-  +  z-  _ 

or  —  —  7^ — 7^  =  1. 


b- 


The  change  from  the  equation  of  the  hyjjerbola  to  that  of 
the  surface  is  effected  by  replacing  z  by  Vy"  +  z"'^. 

x'^      z^ 
3.  The  hyperbola  ^  —  ri  =  !>  Fig.  15,  is  rotated  about  OZ. 

cl  D 

The  equation  of  the  hyperbola  is 

a^           b'^ 
As  Pi  is  rotated  about  OZ,  ON  remains  fixed  and  NPi  becomes 
N  P.     Therefore  

NP'       ON'  ^  -I 


INTERPRETATION  OF  EQUATIONS. 


297 


Fig.  15. 


But  for  the  point  P,  ON  =  z  and  NP  =  Vx''  +  f. 
Hence  the  equation  of  the  surface  is 
jj2  +  y'-^  _  z^  _ 


or 


x^      y'       z- 
a-      a-       b"' 


1. 


The  change  from  the  equation  of  the  hyperbola  to  that  of  the 
surface  is  made  by  replacing  x  by  Vx-  +  y-. 

We  thus  see  that  to  obtain  the  equation  of  a  surface  of  revo- 
hition  formed  by  rotating  a  plane  curve  about  one  of  the  coordi- 
nate axes,  we  leave  the  equation  of  the  plane  curve  unchanged 
as  far  as  the  coordinate  parallel  to  the  axis  of  rotation  is  con- 
cerned, and  replace  the  other  coordinate  by  the  square  root  of 


298  SOLID   ANALYTIC    GEOMETRY. 

the  sum  of  the  squares  of  itself  and  the  coordinate  not  present 
before. 

Conversely,  if  in  any  equation  two  coordinates  appear  only 
in  the  form  of  the  square  root  of  the  sum  of  their  squares  or 
some  algebraic  expression  of  that  quantity,  the  corresponding 
surface  is  one  of  revolution.  The  equation  of  the  generating 
plane  curve  may  be  formed  from  the  given  equation  by  replac- 
ing the  radical  by  either  of  the  coordinates  involved  and 
leaving  the  third  coordinate  unchanged.  The  axis  of  rotation 
is  then  the  axis  of  this  third  coordinate. 

1.  For  example, 

y^  +  z"  =  4px 
is  the  same  as 


(Vy^  +  z-)'''  =  4px, 
and  is,  therefore,  the  equation  of  the  paraboloid  of  revolution 
formed  by  rotating  the  parabola  y^  =  4px  about  OX  as  an  axis. 

2.  x2  +  y2  +  z-  +  2  Gx  +  2  F Vy"-'  +  z-  +  C  =  0 

is  the  same  as 


x2  +  (Vy-'  +  z-)'  +  2  Gx  +  2  F Vy-  +  z-  +  C  =  0 

and  is,  therefore,  the  equation  of  the  surface  formed  l)y  rotat- 
ing the  circle  x^-}-z--|-2Gx-f2  Fz-|-C  =  0  around  OX  as 
an  axis. 

EXAMPLES. 

1.  If  U  and  V  are  expressions  involving  the  coordinates 
X,  y,  and  z,  show  that  I U  +  kV  ^  0  is  the  equation  of  a  sur- 
face passing  through  all  points  common  to  the  surfaces  U  =  0 
and  V  =  0,  and  meeting  them  at  no  other  points. 

2.  If  U  and  V  are  expressions  involving  the  coordinates 
X,  y,  and  z,  show  that  UV  =  0  represents  the  two  surfaces 
U  =  0  and  V  =  0. 

3.  Find  the  equation  of  a  cylinder,  the  base  of  which  is  an 
ellipse  and  the  axis  of  which  is  OX. 


INTERPRETATION   OF    EQUATIONS.  299 

4.  Find  the  equation  of  a  right  circular  cylinder  which  is 
tangent  to  the  plane  XOZ  and  has  its  axis  in  the  plane  ZOY. 

5.  Find  the  equation  of  a  right  circular  cylinder  which  is 
tangent  to  the  planes  XOZ  and  XOY. 

6.  The  axis  of  a  right  circular  cylinder  is  the  line  x  =  a, 
y  =  b,  and  the  radius  of  its  base  is  r.     Find  its  equation. 

What  kind  of  a  surface  is  represented  by  each  of  the  fol- 
lowing equations  ? 

7.  y2-[-z2  +  4y-Gz  =  0. 

8.  z-  +  7x  =  0. 

9.  a V  +  by  —  a-^b^  =  0. 

10.  aV-by-F  a-b2  =  0. 

11.  3x2  _^  (jxy  _^  3y.  _^  2x  +  7  =  0. 

12.  2z-  +  z  — 21  =  0. 

13.  xy  +  2  =  0. 

14.  Find  the  equation  of  the  surface  generated  by  revolv- 
ing the  straight  line  x  =:  3  about  OX. 

15.  Find  the  equation  of  the  cone  generated  by  revolving 
the  line  y  =  mx  about  OX. 

IG.  Find  the  equation  of  the  surface  generated  by  revolv- 
ing the  parabola  y-  =  4px  about  OY. 

17.  Find  the  equation  of  the  surface  generated  by  revolv- 
ing a  straight  line  around  OX  : 

(a)  when  tlie  line  is  perpendicular  to  OX  ; 

(b)  when  the  line  is  parallel  to  OX  ; 

(c)  when  the  line  makes  an  oblique  angle  with  OX. 

18.  Find  the  equation  of  the  ring  surface  formed  by 
revolving  about  OX  the  circle  x-  +  (y  —  b)-  =  a^,  where 
a>  b. 

19.  Find  the  equation  of  the  surface  formed  by  revolving 
ay2  =  x°  about  OX. 

20.  Find  the  equation  of  the  surface  formed  by  revolving 
a^y  =  x"  about  OX. 


300  SOLID   ANALYTIC   GEOMETRY. 

21.  Find  the  equation  of  the  surface  formed  by  revolving 
xi  +  yi  :=  as  about  OX. 

22.  Find  the  equation  of  the  surface  formed  by  revolving 
(x2  +  ff  =  8?  (x2  —  f)  about  OX. 

Interpret  the  following  equations  : 

23.  x-  +  y-  =  al 

24.  z"  +  X-  —  7y  =  0. 

25.  (x^  +  z2)l  +  y^  =  ai 

26.  y*  — 16x^  — 16z-  =  0. 

27.  What  kind  of  a  line  is  represented  by  the  two  equa- 
tions, 3x  —  5  =  0,  4y  +  1  =  0  ? 

28.  What  kind  of  a  line  is  represented  by  the  two  equa- 
tions, x''  -f  z^  -  6y  ==  0,  y  +  2  =  0  ? 

29.  What  kind  of  a  line  is  represented  by  the  two  equa- 
tions, x-  -  9y  -  36  =  0,  X  +  5  =  0  ? 


CHAPTER   III. 

THE    PLANE  AND   THE    STRAIGHT    LINE. 

THE    PLANE. 

16.     Any  Equation  of  the  First  Degree  Represents  a  Plane. 

The  most  general  equation  of  the  first  degree  has  the  form 
Ax  +  By  +  Cz  +  D  =  0.  [10] 

Since,  by  §  12,  this  equation  represents  a  surface,  we  have 
only  to  prove  that  this  surface  is  a  plane. 

Let  Pi(xi,  Yi,  Zi)  and  Po(x2,  y2,  Zg)  be  any  two  points  upon 

the  surface  represented  by  [10].     Then,  since  the  coordinates 

of  a  point  upon  a  locus  satisfy  the  equation  of  the  locus,  we 

have 

Axi+By.  +  Czi  +  D-O,  (1) 

Ax2  +  Byo  +  Czo  +  D  ==  0.  (2) 

Let  us  join  Pj  and  P2  by  a  straight  line,  and  take  a  third 
point  P3,  which  divides  the  distance  PiPo,  internally  or  exter- 
nally, in  the  ratio  of  li  :  l.j.  By  suitably  varying  this  ratio  we 
may  take  P3  any  point  on  the  line  P1P2. 

By  [2],  §  4,  the  coordinates  of  P3  are 

_  I1X2  +  Ui        _  Iiy2  +  l2yi        _  liZo  +  I2Z1 
""'"     I1  +  I2     '^'~     I1+I2    '"'"     Ii  +  l2"' 
If  we  substitute  these  values  in  the  left-hand  side  of  [10], 
it  becomes 

|-^  (Ax2  +  By2  +  Cz2  +  D)  ^-  |-^^ (Axi  +  By,  +  Cz,  +  D). 

This  is  equal  to  zero  by  virtue  of  (1)  and  (2).  Hence  the 
point  Pg,  which  is  any  point  of  the  line  joining  P^  and  Pj, 


302 


SOLID    ANALYTIC    GEOMETKY. 


lies  on  the  surface.  Therefore  the  surface  is  a  plane,  since 
a  straight  line  joining  any  two  of  its  points  lies  entirely 
upon  it. 


17.  To  Find  the  Equation  of  a  Plane  in  Terms  of  the 
Length  and  the  Direction  Cosines  of  the  Normal  from  the 
Origin. 

Z 


Fig.  16. 

Let  ABC  be  the  plane,  OD  the  normal  from  the  origin, 
p  the  length  of  the  normal,  and  cos  a,  cos  ^,  cos  y  its  direc- 
tion cosines.  Take  P  any  point  in  the  plane,  with  the  coor- 
dinates X  —  OL,  y  =  LM,  z  ==  MP.  By  2,  §  6,  the  projection 
of  the  broken  line  OLMP  upon  OD  equals  the  projection  of 
the  straight  line  OP  upon  OD.  The  projection  of  OP  upon 
OD  is  OD  itself,  since  the  angle  PDO  is  a  right  angle.  The 
projection  upon  OD  of  OL  is  x  cos  a,  that  of  LM  is  y  cos  /B, 
and  that  of  M  P  is  z  cos  y.     Hence  we  have 


X  cos  a  +  y  cos  p  +  z  cos  "Y  =  p. 

This  is  called  the  normal  equation  of  the  plane. 
The  general  equation  of  the  plane. 

Ax  +  By  +  Cz  +  D  =  0, 


[11] 


THE    PLANE.  303 

can  be  readily  put  in  the  normal  form  [11  J.  For,  as  it 
stands,  A,  B,  C  are  evidently  proportional  to  cos  a,  cos  /?,  cos  y, 
and,  therefore,  by  §  5, 

A  „  B 

cos  a=  —  ,  cos  B  =  — , , 

Va^+b^+c-  Va^+b^+c^ 

C 

cos  y  =     ,  • 

Va^  +  b^  +  c- 

Hence  the  equation, 

Ax+By  +  Cz+D^,^ 
V  A^'  +  B^  +  C^ 
corresponds  to  [11].  The  ambiguity  in  regard  to  the  sign  of 
the  radical  is  removed  by  noticing  that,  in  [11],  the  absolute 
term,  if  transposed  to  the  left-hand  side  of  the  equation,  is 
negative.  Hence  the  sign  of  the  radical  must  be  so  taken 
as  to  make  the  absolute  term  of  the  general  equation  nega- 
tive. 

18.  To  Find  the  Equation  of  a  Plane  in  Terms  of  its 
Intercepts  upon  the  Three  Axes. 

Let  a,  b,  c  be  the  intercepts  upon  the  axes  of  x,  y,  z,  respec- 
tively. To  find  a,  we  put  y  =  0,  z  =  0  in  [10],  and  take  the 
corresponding  value  of  x.     Hence 

D 

Similarly,  b  =  —  — ' 

D 

Equation  [10]  may  be  written 


_D    ' _D ' _ D 
ABC 

1  +  1  +  1  =  '-  M 


304  SOLID    ANALYTIC    GEOMETRY. 

19.     To  Find  the  Equation  of  a  Plane  which  Passes  through 
Three  Given  Points. 

Since  the  points  are  on  a  plane,  their  coordinates  satisfy 
an  equation  of  the  form 

Ax+  By  +  Cz+  D=rO. 
We  have,  therefore, 

Axi+Byi  +  Czi+D  =  0, 

Ax,+ By2+Cz2+D  =  0, 

Ax3+By3+Cz3+D  =  0. 
These  three  equations  are  in  general  sufficient  to  determine 

ABC 

— ,  — ,  jr,  and  the  values  of  these  quantities  substituted  in  the 

general  equation  give  the  required  result. 
Two  possible  exceptions  deserve  notice. 

ABC 

(1)  It  may  happen  that  the  three  equations  in  — ,  — ,  —  are 

contradictory.      This  would  always  happen,  if  the  true  value 

of  D,  at  the  outset  unknown,  were  zero.     In  this  case  the 

BCD 
student  should  try  other  ratios,  e.g.,  -r,  -r,  -r-. 

(2)  It  may  happen  that  the  three  equations  are  not  inde- 
pendent. There  are,  then,  an  infinite  number  of  solutions, 
and  it  will  be  found  that  the  three  given  points  lie  upon  a 
straight  line. 

20.     Angle  between  Two  Planes. 

Let  the  two  planes  be 

Ax+  By+Cz+  D  =  0, 
and  A'x+B'y  +  C'z+D'  =  0. 

The  angle  between  them  is  equal  to  the  angle  between 
their  normals.     Therefore,  by  [4],  §  7,  if  ^  be  the  angle, 

AA'+BB'+CC 


cos  u^ 


Va-  +  B^  +  C^  VA'2  +  B'2  +  C'^ 


THE    PLANE.  305 

If  the  planes  are  perpendicular, 

AA'+  BB'+CC'  =  0. 

If  the  planes  are  parallel, 

A__B  ^C 
A'~  B'~C'" 

21.  Perpendicular  Distance  from  a  Given  Point  to  a 
Given  Plane. 

Let  the  equation  of  the  given  plane  (M)  be  in  the  normal 
form 

X  cos  a  -[-  y  cos  13  -{-  z  cos  y  =  p. 

Let  Pi  (xj,  yi,  Zi)  be  the  given  point,  and  pi  its  required 
distance  from  the  plane.  Pass  a  plane  (Mi)  through  Pi  par- 
allel to  (M).     Its  equation  will  be 

X  cos  a  +  y  cos  /3  +  z  cos  y  =  p  +  pi, 
when  Pj  is  on  the  opposite  side  of  (M)  from  the  origin,  and 

X  cos  a  +  y  cos  j8  +  z  cos  y  ^  p  —  pi, 
when  Pi  is  on  the  same  side  of  (M)  as  the  origin. 

Since  Pi  is  on  (M,),  we  may  substitute  its  coordinates  and 
obtain 

Xi  cos  a  +  yi  cos  ^  +  Zi  cos  y  =  p  ±  pi, 

or  ±  pi  =  Xi  cos  a  +  Yi  cos  /8  +  Zi  cos  y  —  p. 

Hence  our  working  rule  will  be  : 
Put  the  equation  of  the  plane  in  tJieforni 

X  cos  a  4"  y  cos  i8  +  z  cos  y  —  p  =  0 

and  substitute  for  x,  y,  and  z  the  coordinates  of  the  point.  The 
resulting  value  of  the  left-hand  member  ivill  he  the  distance  of 
the  p)oint  from  the  plane,  and  will  he  jtositive  if  the  p)oint  and 
the  origiii  are  on  opposite  sides  of  the  2'lane,  and  will  he  negative 
if  the  point  and  the  origin  are  on  the  same  side  of  the  plane. 


306  SOLID    ANALYTIC    GEOMETRY. 


THE    STRAIGHT    LINE. 

22.  Any  Two  Simultaneous  Equations  of  the  First  Degree 
Represent  a  Straight  Line. 

For  each  equation,  taken  alone,  represents  a  plane.  The 
values  of  x,  y,  z  which  satisfy  both  equations  represent  points 
which  are  common  to  both  planes.  Such  points  lie  evidently 
upon  the  straight  line  which  is  the  intersection  of  the  two 
planes.  Hence  the  most  general  equations  of  the  straight 
line  are 

Ax  +  By  +  Cz  +  D  =0,  ) 

A'x  +  B'y +  C'z  +  D'  =  0.  i  Li«5J 

These  equations  may  be  put  into  another  form  by  first 
eliminating  z  and  then  y.     There  results 

y  =  qx  +  r,  ( 
z==  sx  +  t.  j 

These  equations  contain  only  four  independent  parameters. 
Therefore  four  conditions  are  sufficient  to  fix  a  straight  line 
in  space.  These  four  conditions  may  be  taken  as  the  coordi- 
nates of  the  points  in  which  the  straight  line  cuts  two  of  the 
coordinate  planes. 


23.  Equations  of  a  Straight  Line  in  Terms  of  its  Direction 
Cosines  and  Any  Known  Point  upon  it. 

Let  Pi  (xi,  yi,  Zj)  be  a  known  point  of  the  line,  P  (x,  y,  z) 
any  point,  and  cos  a,  cos  (3,  cos  y,  the  direction  cosines. 

On  Pi  P  as  a  diagonal  construct  a  parallelopiped,  the  edges 
of  which  are  parallel  to  the  coordinate  axes.     Then 

PiL  =  x  — xi,    LM=y  — yi,   MP  ^z  —  Zj. 


THE    STRAIGHT    LINE. 


307 


P 

y' 

R 

,'' 

/ 

M 

Fig.  17. 

From  the  riglit  triangle  LPiP  we  have 
PiL=  PiPcos  LPiP, 
or  X  —  Xi  ^  r  cos  a, 

if  PiP  is  denoted  by  r. 

Similarly,  y  —  yi  =  r  cos  /3, 

z  —  Zi  :=  r  cos  y. 


Hence 


Xi 


Zi 


[14] 


cos  a         cos  p  cos  "Y 

which  are  the  required  forms  of  the  equations. 

The  following  special  cases  deserve  notice  : 

(1)  One  denominator  zero,  as  cos  a,  for  example.     Omitting, 
for  the  time  being,  the  fraction  containing  this  denominator, 


we  shall  have  left  the  equation 


yi_z  — zi 


,  which  will 


cos  ^        cos  y 

represent  one  plane  tlirough  the  line.     TUit  if  cos  a:=0,  a  is 
some  odd  multiple  of  90°,  and  the  line  is  perpendicular  to 


308  SOLID    ANALYTIC    GEOMETRY. 

OX,  and  is,  accordingly,  in  a  plane  perpendicular  to  OX. 
Since,  however,  one  point  of  the  line  is  in  the  plane  x  =  Xi, 
which  plane  is  perpendicular  to  OX,  the  whole  line  must  be 
in  that  plane.  Therefore,  the  equation  of  the  second  plane 
through  the  line  will  be  x  =  Xi. 

(2)  If  two  denominators  are  zero,  as,  for  example,  cos  a 
and  cos  ji,  the  line,  according  to  case  (1),  will  be  in  the  planes 
X  =  xi  and  y  =  yi,  and  these  equations  will  be  its  equations, 
the  fraction  in  z  being  superfluous.  It  is  evident  that  in  this 
case  the  line  will  be  parallel  to  OZ. 

24.  To  Place  the  General  Equations  of  the  Straight  Line 
in  the  Form  [14], 

Prom  Ax  +  By  +  Cz  +  D  =  0, 1 

A'x+ B'y  +  C'z+D'  =  0,  f  ^^^ 

we  obtain,  as  shown  in  §  22, 

y  =  qx  +  r,)  . 

z  =  sx  +  t.  I  ("'^ 

Equating  the  values  of  x  from  these  two  equations,  we  have 

y  —  r       z  —  t 

X  = = ) 

q  s 

X  —  0        y  —  r        z  —  t 

or  — - —  = = 

1  q  s 

The  coordinates  (0,  r,  t)  correspond  exactly  to  (xi,  yi,  Zi)  of 

[14].     Hence 

cos  a cos  /? cos  y 

1  q  s 

and,  therefore,  by  §  5, 

1 

cos  a  =  —  =^^=  > 

Vl  +  q^  +  s^ 


Vl  +  q^  +s^ 


THE    STRAIGHT    LINE.  309 

Hence  the  equations 

X  —  0         y  —  r         z  —  t 

1  q  s 


Vl  +  q'  +  s^       Vl  +  q'  +  s-       Vl  +  q'  +  s^ 
are  in  the  form  [14]. 

The  above  method  is  to  be  employed  if  the  problem  is  to 
find  the  direction  cosines  of  a  given  line. 

25.  Equations  of  a  Straight  Line  Passing  through  Two 
Given  Points. 

A  line  through  (xi,  yi,  Zi)  has  the  equations 
X  — xi  ^  y  — yi  _  z  — zi  _ 

cos  a  COS  (3  COS  y 

If  it  also  passes  through  (xo,  yo,  Zg),  then 

X2—  Xi  _  y2  —  yi  _  Z2  —  Zi  _ 
cos  a     COS  /3     COS  y 

By  division  we  get,  after  changing  the  signs  of  the  denomi- 
nators, 

xi  —  X2      yi  —  y2      zi  —  Z2  ■-    -■ 

as  the  required  equations  of  the  line. 

The  following  are  the  special  cases  to  be  noted  : 
(1)  One  denominator  zero,  as  Xj  —  Xo,  for  example.     Omit- 
ting, for  the  time  being,  the  fraction  in  x,  we  shall  have  left 
the  equation 

y  —  yi_z  —  zi^ 
yi  —  y2     zi  —  Z2 

which  will  represent  one  plane  through  the  line. 

Now  if  Xi  —  X2  =  0,  X2  =  Xi,  and  two  points  of  the  line  are 
at  the  same  distance  from  the  plane  YOZ,  and  hence  the  line 
is  in  a  plane  parallel  to  the  plane  YOZ  and  distant  Xj  from  it. 
Therefore  the  equation  of  the  second  plane  through  the  line 
will  be  X  =:  Xx. 


310  SOLID    ANALYTIC    GEOMETRY. 

(2)  If  two  denominators  are  zero,  as,  for  example,  Xj  —  Xj 
and  yi  —  ys,  the  line,  according  to  case  (1),  will  be  in  the 
planes  x  =  x^  and  y  =  yj,  and  these  equations  will  be  its 
equations,  the  fraction  in  z  being  superfluous.  The  line  is 
obviously  parallel  to  OZ. 

(3)  If  all  three  denominators  are  zero,  the  two  points  which 
are  to  determine  the  line  are  coincident,  and  hence  no  line  can 
be  determined. 

26.  Plane  through  a  Given  Line  and  Subject  to  One  Other 
Condition. 

If  the  eqixations  of  the  line  are 

Ax+  By +  Cz+  D  =0, 
A'x  +  B'y  +  C'z  +  D'  =  0, 
then  I  (Ax  +  By  +  Cz  +  D)  +  k  (A'x  +  B'y  +  C'z  +  D')  =  0  (1) 
is  the  equation  of  some  plane  through  the  line,  I  and  k  being 
arbitrary  constants.  For  it  represents  some  plane,  since  it  is 
of  the  first  degree ;  moreover,  the  line  lies  entirely  upon  this 
plane,  for  the  coordinates  of  any  point  of  the  line  will  neces- 
sarily satisfy  the  equation  of  the  plane. 

Hence,  (1)  represents  any  plane  passed  through  the  line, 
and  a  second  condition  can  be  imposed  upon  the  plane  by 
giving  different  values  to  I  and  k. 

We  will  illustrate  this  work  by  the  two  following  numerical 
examples. 

Ex.  1.     Find  the  equation  of  a  plane  through  the  line 

x+y +    z=0 
2x  +  y  +  2z  =  l 
and  the  point  (1,  —  1,  1). 

Let  I  (x  +  y  +  z)  +  k  (2x  +  y  +  2z  —  1)  =  0  be  the  required  equation. 
Since  the  plane  passes  through  the  point  (1,  —  1,  1),  the  equation  of  the 
plane  is  satisfied  by  the  coordinates  of  this  point. 

.-.  1(1-1  +  1)  +  k(2-  1  +  2-1)  =  0  or  I  =  -2k. 
.  • .  the  equation  of  the  plane  will  be 

-  2k  (x  +  y  +  z)  +  k  (2x  +  y  +  2z  -  1)  =  0  or  y  +  1  rr  0. 


THE    STRAIGHT    LINE.  311 

Ex.  2.     Find  the  equation  of  a  plane  tlirough  the  line 
x+y +    z=0 

2x  +  y  +  2z=  1 

and  parallel  to  the  plane  3x  +  2y  +  oz  +  2  =  0. 
If  the  required  plane  is 

I  (x  +  y  +  z)  +  k  (2x  +  y  +  2z  -  1)  =  0, 

or  (I  +  2  k)  x  +  (I  +  k)  y  +  (I  +  2  k)  z  -  k  =  0, 

,      .on  I  +  2  k       I  +  k       I  +  2  k       ^   ^  I  +  2  k       I  +  k 

by  }t  20,         — ^ —  —  -  _^  -  =  — - —  .     But  — 3 —  =  -— 

is  the  only  independent  equation  here,  and  from  this  equation  I  =  k. 

.-.  k(x  +  y  +  z)  +  k(2x  +  y +  2z-l)  =  0, 
or       3x  +  2y  +  3z  —  1  =  0      is  the  required  plane. 


EXAMPLES. 

1.  What  is  the  equation  of  a  plane  5  units  distant  from  the 
origin  and  perpendicular  to  the  line  joining  the  origin  to 
(1,  6,  5)  ? 

2.  How  far  is  the  plane  3x  +  4y  —  12z  +  10  =  0  from  the 
origin  ?  Find  the  direction  cosines  of  a  line  perpendicular  to 
this  plane. 

3.  Find  the  equation  of  a  plane  through  the  three  points 
(1,  2,  3),  (—  1,  —  2,  —  3),  and  (4,  —  2,  4).  What  will  be  its 
intercepts  on  the  axes  ? 

4.  What  is  the  angle  between  the  planes 

3x  +  5y  —  Gz  +  1  =  0, 
3x  —  r>y  —  ()z  +  1=0? 

5.  How  far  is  the  point  (1,  5,  0)  from  the  ])lane 

3x-4y  +  12z+26  =  0? 

G.  A  line  making  angles  60°,  45°,  and  C0°,  respectively,  with 
the  axes  of  x,  y,  and  z  passes  througli  point  (1,  —  3,  1).  What 
are  its  equations  ? 

7.  Find  the  equations  of  a  line  passing  through  the  points 
(6,  2,-1)  and  (3,  4,-4). 


312  SOLID    ANALYTIC    GEOMETRY. 

8.  What  are  the  direction  cosines  of  this  last  line  ? 

9.  Find  the  equations  of  a  line  passing  through  the  points 
(1,  5,  —  3)  and  (2,  3,  —  3). 

10.  Find  the  equation  of  a  })lane  passing  through  the  point 
(2,  —  3,  3),  perpendicular  to  the  line  of  Ex.  7. 

11.  Find  the  direction  cosines  of  the  line  of  intersection  of 
the  planes 

3x  +  y  —  6  =  7z, 
2x  —  3y  +  4z  =  7. 

12.  Find  the  equation  of  the  plane  perpendicular  to  the 
line  joining  (2,  4,  6)  and  ( —  6,  —  4,  —  2)  at  its  middle  point. 

13.  Find  the  point  of  intersection  of  the  line 

X— 2_y+l_z— 2 
3      ""      6      ~      1 
with  the  plane  3x  +  4y  —  6z  =  0. 

14.  Find  the  angle  which  the  line 
3x+2y  +  7  =  0 
2y  +  4z+3  =  0 

makes  with  the  plane  x-l-y  +  3z  —  2^=0. 

15.  Find  the  equation  of  the  locus  of  points  at  a  distance 
+  3  from  the  plane  x  +  y  +  zH-3  =  0. 

16.  Find  the  locus  of  points  equally  distant  from  the 
planes 

x  +  2y  +  3z  +  4  =  0, 
x-2y-f  3z-5=0. 

17.  A  plane  is  passed  through  (2,  6,  —  7)  parallel  to  the 
plane  3x  —  4y  +  12z  +  17  =  0.  Find  its  equation  and  dis- 
tance from  the  given  plane. 

18.  Determine  the  equation  of  a  plane  through  (1,  —  3,  1) 
and  (3,-1,  3)  which  shall  be  perpendicular  to  the  plane 
2x-3y  +  Gz  — 14  =  0. 

19.  What  are  the  equations  of  a  line  through  (1,  3,  —  5) 
parallel  to  the  line 

y  =  3x-14^^ 
7x-2z  =  17  \  ' 


THE    STRAIGHT    LINE.  313 

20.  Is  the  point  (6,  2, 1)  on  the  straight  line  determined  by 
(2,  4,  3)  and  (-  1,  6,  -  1)  ? 

21.  Find  the  angle  between  the  lines 


2  =  3y-2j  (z  =  2x  +  3 

22.  If  the  above  lines  meet,  find  their  point  of  intersection. 

23.  Where  is  the  point  of  intersection  of  the  lines 

3x  — y  — 3z  — 8  =  0)        ,(     x  +  y  — z  =  0 
'  and  '  -' 


X  — y+    z  +  2  =  0j  |6x  — Oy  — 3z  — lo  =  0j' 

24.  What  is  the  equation  of  the  plane  determined  by  the 
point  (1,  3,  —  1)  and  the  line 

2x  +  r>y  -  6  =  0 
3y  +  7z  +  1  =  0 

25.  What  is  the  equation  of  the  plane  determined  by  the 
two  lines 

S«-    y_l3  =  0)  <7x-;)y-22  =  0) 


y  +  4z+    l  =  0S'""M2x  +  3z-    3  =  oi- 

20.   Find  the  equation  of  a  plane  passing  tlirough  the  line 
I  X  +  2y  +  1  =:  0  ( 
j  y-3z  +  4  =  0  j 
and  perpendicular  to  the  plane  x  +  y  +  z  =  0. 

27.  Find  tlie  equations  of  the  projection  on  the  plane 
2x  +  6y  —  3z  +  7  =  0  of  the  line 

4x  -  3y  +  15  =  0  I 
3y  +    z  -    1  =  0  j  • 

28.  Prove  that  the  three  planes 

x  +  2y  -  3z  +  1  =  0, 
y  +  5z-l  =  0, 
3x  +  4y  — 19z  +  5  =  0 
pass  through  the  same  straight  line. 

29.  A  line  is  drawn  through  (1,  2,  3)  in  the  direction 
a  =  45°,  p  =  60°,  y  =  60°.  Find  the  coordinates  of  a  point 
upon  this  line  distant  5  units  from  (1,  2,  3). 


314  SOLID    ANALYTIC    GEOMETRY. 

30.  Find  a  point  in  the  plane  XOY  equally  distant  from 
the  origin  and  the  points  (1,  2,  —  1)  and  (3,  —  1,  —  3). 

31.  Find  a  point  in  the  plane  x  +  y  +  z  =0  equally  distant 
from  the  points  (0,  1,  0),  (1,  —  2,  1),  and  (2,  3,  1). 

32.  Find  the  centre  of  a  sphere  of  radius  7  passing 
through  (—  3,  1,  —  9)  and  (1,  —  5,  3)  and  tangent  to  the 
plane  2x  +  3y  +  6z  -  23  =  0. 

33.  Find  the   distance  of  the  point  (4,  5,  —  6)  from  the 

line  (3x-2y  +  7  =  0 

(  2y  +  z  -  3  =  0. 

34.  Find  the  distance  between  the  lines  determined  by  the 
points  (1,  2,  1),  (- 1,  2,  -  1),  and  (2,  1,  3),  (3,  1,  2). 

35.  Show  that,  if  the  lines 


qx  +  r  )  (  y  =  q'x  +  r' 

^  '  and  ' 


intersect, 


z  =  sx  +  't  )  (  z  =  s'x  +  t' 

r— r'        t  — t' 


q  —  q        s  —  s 

36.  If  cos  ai,  cos  ySi,  cos  y^  and  cos  ag,  cos  p^,  cos  yo  are  the 
respective  direction  cosines  of  any  two  lines,  and  cos  a,  cos  yS, 
cos  y  are  the  direction  cosines  of  a  line  perpendicular  to  the 
two  given  lines,  show  that 

cos  a cos  p 

cos  )8i  cos  y2  —  cos  ^2.   cos  yi   cos  yi  COS  ag  —  COS  yg  COS  a^ 

COS  y 


cos  y 
COS  ai  COS  182  —  COS  a-i  COS  /?! 


37.  A  point  moves  so  that  the  sum  of  its  distances  from 
any  number  of  given  planes  is  constant.  Show  that  its  locus 
is  a  plane. 

38.  Pi  (xi,  yi,  zi)  and  P2  (xo,  y^,  z.,)  are  any  two  given 
points.  Find  the  locus  of  a  point  Q  of  the  plane  XOY  so  that 
the  angle  P1QP2  is  a  right  angle. 


THE   STRAIGHT   LINE.  315 

39.  If  Pi  (xi,  yi,  zj)  and  P2  (xa,  y2,  Zo)  are  any  two  given 
points,  find  the  locus  of  a  third  point  Q  snch  that  PjQ  and 
P2Q  are  equal,  or  are  in  a  constant  ratio. 

40.  Show  analytically  that  the  locus  of  points  equally  dis- 
tant from  any  three  given  points  is  a  straight  line  perpendic- 
ular to  the  plane  of  the  three  points. 

(Take  the  three  points  in  the  plane  z  =  0.) 

41.  Show  analytically  that  the  planes  bisecting  the  diedral 
angles  of  any  triedral  angle  meet  in  a  common  line. 

42.  If  the  normal  distance  from  the  origin  of  the  plane 
which  makes  the  intercepts  a,  b,  and  c  respectively  on  the 
axes  of  X,  y,  and  z  is  p,  prove  that 

p2      a^  ^  b^  ^  c' 

43.  If  A,  B,  and  C  of  equation  [10]  approach  the  limit 
zero,  show  that  the  limiting  form  of  the  equation  is  D  =  0 
and  that  the  plane  becomes  infinitely  distant  from  the  origin. 

44.  What  two  planes  are  represented  by  the  equation 

x^  +  4xy  -f  3y^  +  xz  +  yz  -f  X  +  5y  +  2z  —  2  =  0? 


CHAPTER  IV. 

EQUATIONS  OF  THE  SECOND  DEGREE. 


27.     The  Cones :  -,  ±  ^  ±  -^  =  0. 
a-      D-      c^ 

If  we  take  all  the  different  combinations  of  the  algebraic 
signs  possible,  we  see  that  there  are  four  surfaces,  the  equa- 
tions of  which  are  respectively 

"^-f.  4-^  =  0      "^-t-L^^o 
a^      b2"^c-^        '    a^      b^      c^' 

But  the  last  three  equations  agree  in  having  two  terms  of 
the  same  sign,  the  third  term  being  of  the  opposite  sign,  so 
that  we  have  essentially  but  two  equations  to  discuss,  i.e., 

i!+i;.+L;=o, 

a^       b-       c^ 

2  9  9 

and  ^  +  ^  -  ^  =  0. 

a^      b-      c" 

a^  ^  b'^  ^  c' 
This  equation  can  be  satisfied  in  real  coordinates  by  (0,  0,  0) 
only,  for  it  may  be  written  (-]  +(f)  +('")  ^=^j  '^^^^  ^^^^ 

sum  of  three  squares  cannot  be  zero  unless  each  of  them  is 
zero.  Therefore,  this  equation  represents  a  single  real  point, — 
the  origin.      If  imaginary  values  of  the  coordinates  are  con- 


EQUATIONS    OF    SECOND    DEGREE.  317 

sidered,  we  should  say  that  the  equation  represents  an  imagi- 
nary cone,  the  proof  being  carried  on  as  in  II. 

„.  ?_: + ^:  _  ?!, = 0. 

a-      b-       c- 

It  is  evident  that  the  origin  is  a  point  of  this  surface,  since 
the  equation  is  satisfied  by  (0,  0,  0).  Moreover,  if  (xj,  y^,  Zj) 
is  a  point  of  this  surface,  ( —  Xj,  —  yi,  —  Zj)  will  also  satisfy 
the  equation,  and  the  surface  is  symmetrical  about  the  origin 
as  a  centre.  In  fact,  this  surface  is  a  cone  with  its  vertex  at 
the  origin.  For,  if  P^  (xi,  yi,  Zj)  is  any  point  of  this  surface, 
the  coordinates  of  any  point  on  the  straight  line  through  the 
origin  and  P^  are  mxi,  myi,  mzi,  where  m  is  an  abritrary 
factor.     Substituting  these  coordinates   in  the  equation,  we 

which  is  true,  since  Pi  (xj,  yi,  Zj)  is  on  the  surface.  Hence 
the  line  through  the  origin  and  Pi  lies  entirely  upon  the  sur- 
face, so  that  the  surface  is  composed  of  straight  lines  passing 
through  the  origin  and  is,  accordingly,  a  cone. 

We  will  complete  the  study  of  this  surface  by  the  method 
of  sections  made  by  planes  parallel  to  the  coordinate  planes. 

If  we  let  z  =  0,  the  resulting  equation  in  x  and  y  will  be 

the  equation  of  the  plane  section  of  the  cone  made  by  the 

plane  XOY,  of  which  the  equation  is  z  =  0.     This  equation  is 

x"       f 

~2~t"  u2^^^'  ^°  ^^^^^  ^^^^^  section  is,  by  §  71,  Part  I,  a  single 

di  D 

point,  the  origin. 

If  we  let  z  =  Zi,  some  fixed  finite  value,  the  resulting  equa- 
tion in  X  and  y  will  be  the  equation  of  the  plane  section  made 
by  the  plane  z  =  Zi.  Por,  by  placing  z  z=  Zi  instead  of  z  =  0, 
we  have  virtually  transferred  the  plane  XOY,  parallel  to  itself, 

X^  y2         2,^ 

through  the  distance  z,.     This  equation  is— „-l-fT, r^O, 

a"*      b"       c" 


318  SOLID    ANALYTIC    GEOMETRY. 


which  may  be  written  — —,  +  /^  \2  —  1-    Hence  the  sec- 


c   7        y  c 

tion  made  by  a  plane  parallel  to  the  plane  XOY  is  an  ellipse, 
which  increases  in  magnitude  as  the  cutting  planes  are  taken 
farther  away  from  the  origin,  since  its  semi-axes  are 

"-^and^^\  . 

c  c 

If  z  =  —  Zi,  we  shall  get  a  section  equal  to  that  made  by 
the  plane  z  =  Zi,  so  that  sections  made  by  planes  parallel  to 
the  plane  XOY,  equally  distant  from  the  origin  but  on  oppo- 
site sides  of  it,  will  be  equal.  Hence  the  cone  is  symmet- 
rical with  respect  to  the  plane  XOY. 

If  y  ==  0,  the  equation  of  the  section  made  by  the  plane 

x^      z^ 
XOZ  is  —^ 2  ^  ^'  which  may  be  written 

a      c  /  ya      c y 
Therefore,  by  §  71,  Part  I,  this  section  is  the  pair  of  straight 

X         Z  X         z 

lines, =  0  and  — I —  =  0,  which  intersect  at  the  origin. 

a       c  a       c 

If  y  =  Yi,  the  resulting  equation  is  —  -(-  ^-, ;,  =  0?  which 

a"       b"      c" 

z-  x^ 

may  be  written  — — : ; — :  =  1.      Hence  the  section 

^cyiV       /ayi-  ^ 

b;         (,b 

made  by  a  plane  parallel  to  the  plane  XOZ  is  an  hyperbola 

having  its  transverse  axis  parallel  to  OZ. 

Since  y  =  —  yi  gives  an  equal  section,  the  cone  is  sym- 
metrical with  res2:)ect  to  the  plane  XOZ. 

As  X  and  y  appear  symmetrically  in  the  equation,  it  follows 
from  the  above  that  the  plane  YOZ  cuts  the  cone  in  a  pair  of 
straight  lines  intersecting  at  the  origin,  and  planes  parallel 
to  the  plane  YOZ   cut  the  cone  in  hyperbolas  having  their 


EQUATIONS    OF    SECOND    DEGREE. 


319 


transverse   axes    parallel  to  OZ,   and  also  that  the  cone  is 
symmetrical  with  respect  to  tlie  plane  YOZ. 

We  may  now  sum  up  our  work  as  follows  :  TJt.is  surface  is 
a  cone  generated  by  a  stra'ujht  line  passhuj  throwjli  the  origin 
and  following  an  ellipse  as  directrix*  The  term  which  has 
its  algebraic  sign  unique  will  show  which  of  the  coordinate 
axes  is  the  axis  of  the  cone. 


Fig.  18. 


If  b^  =  a",  the  equation  becomes 
have  a  circular  cone. 


x^  +  y^ 


-„  ^  0,  and  we 


*  This  theorem  might  have  been  stated  immediately  after  the  finding 
of  the  first  elliptic  section,  but  we  continued  to  investigate  other  sections 
because  of  the  instructiveness  of  tlie  method. 


320  SOLID   ANALYTIC    GEOMETRY. 

X-      y^      z^      , 
28.     The  Central  Quadrics :  ^  ±  :j^  ±  ^^  =  1. 

The  distinct  types  of  equation  here  are 

^  _  y!  _  l' = 1 

a-2  b2  ^,2-1- 

In  the  case  of  each  of  these  equations  it  is  evident  that  if  it 
is  satisfied  by  (xi,  yi,  Zi)  it  is  also  satisfied  by  (—  Xi,  —  yi,  —  Zj). 
Hence  the  origin  is  a  centre  of  each  of  these  surfaces,  for 
which  reason  they  are  called  central  quadrics. 

Proceeding  to  a  separate  discussion  of  each  of  these  sur- 
faces we  will  take  them  up  in  the  order  in  which  their  equa- 
tions are  written  above. 

a-      b-      c- 

If  z  =  0,  the  equation  of  the  section  made  by  the  plane 

x^      v^ 
XOY  is   -  +  ;-:;  =  l.     This  section,  then,  is  an  ellipse  with 
a'^      b" 

semi-axes  a  and  b. 

If  z  =  Zi,  the  resulting  equation  is 

3^2-1-1,2+^2       -^o^a^+b-i  c^' 

and  there  are  three  cases  to  consider  : 

(1)  If  Zi  <  c,  1 ^  is  positive,  and  the  equation  may  be 

written  x^  , y^ ^ 


(w-^y  (w-«' 


This  section  is,  accordingly,  an  ellipse  which  decreases  in 

magnitude  as  z^  increases  toward  c  as  a  limit. 

x^       y2 
(2)  If  Zi  =:=  c,  the  equation  becomes  "i  +  r^  ==  0,  and  the 

section  is  a  single  point  on  the  axis  of  z. 


EQUATIONS    OF    SECOND    DEGREE.  321 

(3)  If    Zi  >  c,   1 ^    is    negative,    and    the    section  is 

imaginary. 

Since  z  =  —  zj  gives  an  equal  section  to  tliat  made  by  tlie 
plane  z  =  z^,  the  surface  is  symmetrical  with  respect  to  the 
plane  XOY. 

Therefore,  this  surface  is  symmetrical  with  respect  to  the 
plane  XOY,  is  bounded  by  the  planes  z  ^  —  c  and  z  =  c,  and 
all  sections  made  by  planes  parallel  to  the  plane  XOY  are 
ellipses,  which  decrease  in  magnitude  as  we  go  farther  from 
the  coordinate  plane  and  finally  reduce  to  the  two  points 
(0,  0,  —  c)  and  (0,  0,  c). 

Since  x,  y,  and  z  appear  symmetrically  in  the  equation  we 
can  say  immediately  that  : 

(1)  The  surface  is  symmetrical  with  respect  to  the  plane 

XOZ,  is  bounded  by  the  planes  y  =  — b  and  y  =  b,  and  all 

sections    made   by   planes    parallel    to    the   plane    XOZ    are 

x^       z^ 
ellipses,  of  which  — ^  -| — ^  =  1   is   the  largest,  and  which  de- 
a        c 

crease  in  magnitude,  as  we  go  from  the  coordinate  plane,  to 

the  points  (0,  —  b,  0)  and  (0,  b,  0). 

(2)  The  surface  is  symmetrical  with  respect  to  the  plane 
YOZ,  is  bounded  by  the  planes  x  =  —  a  and  x  =  a,  and  all 
sections  made  by  planes  parallel  to  the  plane  YOZ  are  ellipses, 

of   which  f  2  -f  -„  =  1  is  the  largest,  and  which  decrease  in 
^  b       c- 

magnitude,  as  we  go  from  the  coordinate  plane,  to  the  points 

(—  a,  0,  0)  and  (a,  0,  0). 

This  surface  is  an  ellipsoid,  of  semi-axes  a,  b,  and  c,  and  is 

represented  in  Fig.  19  for  the  case  in  whicli  a  >  b  >  c. 

x^      y^  +  z^ 
If  c  =  b,  the  equation  becomes  —  +  ^— r-^ —  =  1,  the  equa- 
tion of  the  ellipsoid  of  revolution,  called  the  prolate  spheroid, 
formed  by  revolving  an  ellipse  around  its  major  axis,  OX. 


322 


SOLID    ANALYTIC    GEOMETllY. 


Fig.  19, 

If  b  =  a,  the  equation  becomes  ~-  +  —  =  1,  the  equa- 

d,  c 

tion  of  the  ellipsoid  of  revolution,  called  the  oblate  spheroid, 

formed  by  revolving  an  ellipse  around  its  minor  axis,  OZ. 

x^  +  y^  +  z^ 

If   c  =  b  ^  a,   the    equation    becomes   —^ =  1,  or 

a 

x"  -j-  y^  +  z^  ^  a^,  which  is  the  equation  of  the  sphere. 
II. 


^"  .  y_  _  ^  =,  1 

a-  ^  b^      c^ 


If  z  ^  0,  the  equation  of  the  section  made  by  the  plane 

x^       y2 
XOY  is  -;;  +  r^=:l,  so  that  this  section  is  an  ellipse  with 
a''      b^ 

semi-axes  a  and  b. 

If  z  :=  Zi,  the  resulting  equation  is 


a^~^  b^ 


=  1  or 


{^^^  (^V 


C" 


=  1. 


Hence  the  section  made  by  a  plane  parallel  to  tlie  plane 
XOY  is  an  ellipse,  which  increases  in  magnitude  without 
limit  as  we  go  from  the  coordinate  plane. 

Since  z  :=  —  Zj  gives  an  equal  section,  the  surface  is  sym- 
metrical with  respect  to  the  plane  XOY. 


EQUATIONS    OF    SECOND    DEGREE  323 

If  y  ^  0,  the  equation  of  the  section  made  by  the  plane 

XOZ  is  — ^  =  1,  so  that  this  section  is  an  hyperbola  with 

dL        c 

its  transverse  axis  on  OX. 

If  y  =  y,,   the    resulting   equation   is   —,-\-Tr,  — -^  =^  Ij  or 

a''       b-       c- 

-^ z=^l  —  7^,  and  there  will  be  three  cases  to  consider  : 

a^       c^  b- 

y,^ 

(1)  If  Yi  <  b,   1  —  r-^  is  positive,  and  we  may  write  the 

equation  x'^  _  z-  

— r,\  2        ^^ 

b^ 

which  is  the  equation  of  an  hyperbola  with  its  transverse 
axis  parallel  to  OX. 

(2)  If  Yi  ^  b,  the  equation  becomes 


'Vi-0  (Wi- 


which  is  tlie  equation  of  a  pair  of  intersecting  straight  lines. 

(3)  If 
equation 


(3)  If  yi  >  b,  1  —  =j-^,  is  negative,  and  we  may  write  the 


(W^-0^  (^V? 


^-1 


1, 


which  is  the  equation  of  an  hyperbola  with  its  transverse  axis 
parallel  to  OZ. 

Examining  these  three  cases,  we  see  that  as  y^  increases 
from  0,  the  vertices  of  the  hyperbola  come  nearer  together, 
till,  when  y^  =  b,  they  coincide  and  the  hyperbola  reduces  to 
a  pair  of  intersecting  straight  lines,  and  that  when  yi  >  b,  tlie 
axis  of  the  hyperbola  is  turned  through  an  angle  of  90°,  and 
the  vertices  go  farther  apart  as  y^  increases. 

Since  y  =^  —  yj  gives  an  equal  section,  the  surface  is 
symmetrical  with  respect  to  the  plane  XOZ. 


324  SOLID    ANALYTIC    GEOMETRY. 

As  X  and  y  appear  symmetrically  in  the  equation,  it  is  evi- 
dent that  the  surface  is  symmetrical  with  respect  to  the  plane 
YOZ,  and  the  plane  sections  made  by  planes  parallel  to  the 
plane  YOZ  are  :  (1)  hyperbolas  with  the  transverse  axis  par- 
allel to  OY,  if  the  cutting  plane  is  between  the  planes  x  =  —  a 
and  X  ==  a  ;  (2)  a  pair  of  intersecting  straight  lines  for  the 
planes  x  =  —  a  and  x  =  a  ;  (3)  hyperbolas  with  the  trans- 
verse axis  parallel  to  OZ,  if  the  cutting  plane  is  outside  the 
space  bounded  by  the  planes  x  ^  —  a  and  x  ^  a. 

In  the  above  work  we  have  seen  that  XOZ  and  YOZ,  both 
of  which  pass  through  OZ,  cut  this  surface  in  hyperbolas. 
The  question  naturally  arises  :  Do  all  the  planes  through  OZ 
cut  this  surface  in  hyperbolas  ? 

If  we  rotate  the  axes  about  OZ  as  an  axis  through  any 
angle  $,  the  formulas  of  transformation  will  be 

X  :=  x'  cos  0  —  y'  siu  9, 
y  ^=x'  sin  ^  +  y'  cos  6, 
z  =  z'; 
for  in  [8],  §  9, 

ai^e,  ^1  =  90°-^,  yi  =  0°, 

a2  =  90°  +  ^,  /82  =  0,  72  =  0°, 

as  =  90°,          /?3  =  90°,         73  =  0°. 

The  equation  of  the  surface  becomes 
'cos^  0  ,  sin^  0 


J  +  2x'y'  cos  6  sin  d  f—^ i  ) 


,     ,,  /sin^  6  .  cos^  ^\       r        . 

+  y'(  -T^  +  ^T^  )--2  =  l• 
If  y':=0,  the  equation  of  the  section  made  by  the  plane 
X'OZ'  is 

""    '  ^^  +  ~b^  J  ~  ^  ~    ' 


EQUATIONS   OF   SECOND   DEGREE. 


325 


or 


V 


F^' 


which  is  the  equation  of  au  hyperbola. 

But  as  0  is  given  different  values,  the  plane  X'OZ'  becomes 
any  plane  through  01,  and,  therefore,  all  sections  of  this 
surface  made  by  planes  through  OZ  are  hyperbolas. 

The  surface  is  an  unparted  hyperboloid  or  hyperboloid  of 
one  sheet,  with  OZ  as  its  axis,  and  is  represented  in  Fig.  20 
for  the  case  in  which  a  >  b  >  c. 


Fig.  20. 


If  b  =:  a,  the  hyperboloid  becomes  one  of  revolution,  OZ 
being  the  axis  of  revolution,  and  its  equation  is 

x'  +  y-  _L=:i^ 


III. 


y 

b- 


326  SOLID    ANALYTIC    GEOMETRY. 

If  X  =  0,  the  equation  of  the  section  made  by  the  plane 
YOZ  is  f,_;  H — ^  =  —  1,  so  that  this  section  is  imaginary. 
If  X  =  Xi,  the  resulting  equation  is 


xi^  _  r  _  z 

a-       b^       c 
and  there  will  be  three  cases  to  consider 


a^       b^       c?  ^      °''  b^  "^  c^  ~  a^         ' 


(1)  If  Xi  <C  a,  — V  —  1  is  negative  and  the  section  is  imagi- 
nary. 

(2)  If  Xi  ^  a,  the  equation   becomes  ^  -| — ^  ^  0,  and  we 
have  a  single  point  on  the  axis  of  x. 

(3)  If  Xi  >  a,  -^  —  1  is  positive  and  the  equation  may  be 

3." 

written  .    v^  z^ 


(WS 


\    ^2 


c 
a 


Therefore,  all  these  sections  will  be  ellipses  of  increasing 
magnitude  as  the  cutting  planes  are  taken  farther  from  the 
coordinate  plane. 

Since  x  ^  —  Xx  gives  an  equal  section,  the  surface  is  sym- 
metrical with  respect  to  the  plane  YOZ. 

Therefore  this  surface,  which  is  symmetrical  with  respect 
to  the  plane  YOZ,  cuts  OX  at  the  points  (—  a,  0,  0)  and 
(a,  0,  0)  and  has  no  points  between  the  planes  x  =  —  a  and 
X  =  a,  while  outside  the  space  bounded  by  these  planes  it 
increases  in  magnitude  as  we  go  from  the  coordinate  plane, 
all  the  sections  parallel  to  the  plane  YOZ  being  ellipses. 

If  z  ^  0,  the  equation  of  the   section   made  by  the  plane 

X^  y2 

XOY  is  —  —  ^  ==  1,  so  that  this  section  is  an  hyperbola  with 
a        D 

its  transverse  axis  on  OX. 


EQUATIONS    or    SECOND   DEGREE. 


327 


If  z  =  Zi,  the  resulting  equation  may  be  written 


b"  c- 


(aVl  +  f)'      (bVl  +  l') 


i  =  l- 


Therefore,  the  section  made  by  any  plane  parallel  to  the 
plane  XOY  is  an  hyperbola  with  its  transverse  axis  parallel 
to  OX. 

The  surface  is  also  symmetrical  with  respect  to  the  plane 
XOY,  for  z  =  —  Zi  and  z  =  Zi  give  the  same  result. 

Similarly,  the  surface  is  symmetrical  with  respect  to  the 
plane  XOZ,  all  sections  parallel  to  that  plane  being  hyper- 
bolas with  the  transverse  axis  parallel  to  OX. 

It  may  also  be  shown  that  every  plane  through  OX  cuts 
this  surface  in  an  hyperbola  with  its  transverse  axis  on  OX. 

The  surface  is  a  biparted  hyperboloid,  or  hyperboloid  of 
two  sheets,  having  OX  as  its  axis  (the  axis  always  being 
determined  by  the  term  of  unique  sign),  and  consists  of  two 
separate  parts  extending  indefinitely  away  from  the  planes 
X  ==  —  a  and  x  =  a. 

Fig.  21  is  for  the  case  when  a  >  b  >  c. 


-f-r-x 


Fig.  21. 

If  c  =  b,  we  have  the  hyperboloid  of  revolution, 

a^  b^  ""    ' 

of  which  the  axis  of  revolution  is  OX. 


328  SOLID    ANALYTIC    GEOMETRY. 


29.     Relation  between  the  Cone  r^  +  vi "~  ll-  =  0  and  the 

HOC 

Hyperboloid  a^  +  v-  ~  c^  ~ 

If  in  place  of  x,  y,  and  z  Ave  use  r,  cos  a,  cos  ji,  cos  y  as  the 
coordinates  of  a  point,  the  equation  of  the  cone  becomes 

COS"  a        COS^  P  _  cos-   ;-   x  _  ., 

'^T  '^     b^          ~^  )  ~  ^-'■^ 

and  that  of  the  hyperboloid  becomes 

COS^  a        COS^  ^        COS^  y"     

a-  b^  c" 


1  COS^  a        COS-  yS        COS-  y 


(2) 


If  now  r,  COS  a,  cos  /?,  and  cos  y  are  the  coordinates  of  a 
point  coanmon  to  the  two  surfaces,  r  cannot  be  zero  since  the 
hyperboloid  does  not  pass  through  the  origin.  Therefore  (1) 
cannot  be  satisfied  unless 

COS^  a        COS^  /3        cos-  y 

a-  b  c 

But  then  we  have  from  (2)  -  =  0,  whence  r  =  oo. 

r 

Therefore  the  cone,  i.e.,  any  element  of  the  cone,  intersects 

the  hyperboloid  at  infinity. 

x"       y"       z^ 
Hence  we  say  that  -^-\-  tt^ 2  ^^  ^  '*'*  ^^*^  asym/ptotic  cone  of 

the  hyperboloid 

x"       y-       z" 
In  the  same  way  we  may  prove  —,  —  ^ 2  ^=  ^  to  be  the 

o  9  9 

X        y        z 
asymptotic  cone  of  the  hyperboloid  —^  —  f^ ^  ^^  1- 


EQUATIONS    OF    SECOND   DEGREE.  329 

30.     The  Unparted  Hyperboloid  is  a  Ruled  Surface. 

If  we  put  the  term  ~  on  the  right-hand  side  of  the  equa- 
tion, the  equation  of  this  hyperboloid  becomes 

9  9  o 

^  _  £  _  1  _  r 

which  may  be  written  in  the  form 

since  each  member  is  the  difference  between  two  squares. 

Now,  if  ki  is  any  constant  whatever,  it  is  evident  that  the 
above  equation  may  be  regarded  as  the  product  of  the  two 
equations  /         w 

a       c  y         b 

a       c       ki  I  b 

But  these  last  two  equations,  being  simultaneous  equations 
of  the  first  degree,  are  the  equations  of  a  straight  line.  This 
line  lies  entirely  upon  the  surface  of  the  hyperboloid,  for  the 
coordinates  of  any  point  which  satisfy  these  equations  will 
necessarily  satisfy  the  equation  of  the  hyperboloid.  By 
giving  ki  different  values  we  can  get  a  whole  system  of  lines 
lying  entirely  upon  the  surface  of  the  hyperboloid.  The  par- 
ticular line  of  the  system,  which  passes  through  any  given 
point  (xi,  Yi,  Zj),  is  found  by  substituting  Xj,  yj,  Zi  in  either 
equation  and  thus  determining  k^  Now,  a  surface  which  may 
be  generated  by  means  of  a  straight  line  is  called  a  ruled 
surface,  the  sim})lest  examples  being  the  cone  and  the  cylin- 
der. It  appears,  therefore,  that  the  iinjjarted  hyperboloid  is  a 
mded  surface. 

There  is  one  other  way  in  which  the  factors  of  the  two 
members  of  the  equation 


330  SOLID    ANALYTIC    GEOMETRY. 


a-c     a  +  n  =  0-n  0  +  ^B 


i_£  =  k /i  +  y 


may  be  paired,  i.e., 


c         "  V     ^ 

These  equations  represent  a  second  system  of  straight  lines 
lying  entirely  upon  the  surface  of  the  hyperboloid.  Hence 
it  appears  that  through  any  point  of  the  unparted  hyperbo- 
loid, two  straight  lines,  one  of  each  of  the  above  systems, 
may  be  drawn  upon  the  surface.  These  lines  are  called  the 
rectilinear  generators,  and  every  point  of  the  surface  may  be 
regarded  as  being  determined  by  the  intersection  of  two  of 
them. 


X  V" 

31.  The  Paraboloids :  T2^u  —  4pz. 

There  are  bvit  two  surfaces  to  be  considered  here  :  (1)  the 

'>  2 

X  V 

elliptic  paraboloid,  -^  +  ^  =  4pz,  represented  in  Fig.  22  for 

the  case  in  which  a  >  b  and  p  >  0  ;  (2)  tlie  hyperbolic  parab- 

oloid,  —  —  7^  =  4pz,  represented  in  Fig.  23  for  the  case  m 
a'       b- 

which  a  >  b  and  p  !>  0. 

The   detailed  study  of  these   surfaces   by  the   method  of 

plane  sections  is  left  to  the  student,  the  work  of  §  28  serving 

as  a  guide. 

32.  It  will  be  noticed  that  the  equations  of  the  cones  and 
the  central  quadrics  can  all  be  represented  by  the  single  equa- 
tion ^  +  fi  +  ~2  =  K  where   k  is   either  zero  or  unity,  and 

a        b        c 

where  a'^,  b^.  and  c^  can  be  given  either  sign,  according  to  the 


EQUATIONS    OF    SECOND    DEGREE. 
Z 


331 


Fig.  22. 

surface  we  wish  to  study.  Accordingly,  we  will  use  this 
equation  in  the  work  that  follows,  writing  our  formulas  as 
general  formulas  in  which  the  student  must  make  the  ap- 
propriate substitutions  for  a^,  b^,  c^,  and  k  in  any  particular 
problem. 


Fig.  23. 


332  SOLID    ANALYTIC    GEOMETKY. 

33.     Tangent  Plane. 

The  locus  of  the  taiujent  lines  drawn  to  the  surface  at  any 
'point  is  a  plane  called  the  tangent  plane,  the  point  being  called 
the  point  of  tangency  or  the  point  of  contact. 

Let  Pi  (xi,  yi,  Zi)  be  any  point  of  the  surface 

x^      v^      z^ 

a"      b-       c''  ^ 

Then,  ^1j^Y1^^1  =  ^,  (2) 

Transforming  to  P^  as  a  new  origin  by  formulas  [7],  §  8, 
we  get,  as  the  equation  of  the  surface, 

^'_l/'_l^_!  .  2xix_'      2y^y'      2ziz'      x^^      y,=^      zi_^_, 
a^"^  b^'"^  c^'"^    a2    +-b2    "^     c^    "^a^^  b=^"^  c^~    ' 

x'^      y'^      z"      2xix'      2yiy'      2ziz' 

by  reason  of  (2). 

If,  now,  we  let  x'  =  r'  cos  a,  y'  =  r'  cos  y8,  and  z'  =  r'  cos  y, 
(3)  becomes 

,2 /COS^  a    .    COS"  y8        COS^  y\ 

+  2rY^^"  +  ^^^  +  ^^)=0,  (4) 

in  which  the  values  of  r'  are  the  distances  from  P^,  the  new 
origin,  to  the  surface  measured  along  the  line  having  as  its 
direction  cosines  cos  a,  cos  /?,  cos  y. 

One  value  of  r'  is  always  zero,  since  the  origin  is  a  point  of 
the  surface.  And  if  cos  a,  cos  /?,  cos  y  have  such  values  that 
this  line  is  any  one  of  the  tangent  lines  to  the  surface,  both 


EQUATIONS    OF    SECOND   DEGREE.  333 

values  of  r'  in  the  above  equation  must  be  zero,  and  this  can 
happen  only  when 

Xi  cos  g      yi  cos  /3      zi  cos  y  _ 

a-  b''  c"  ^  '' 

Multiplying  (5)  by  r'  and  then  replacing  r'  cos  a  by  x',  etc., 
we  have  ,  ,  , 

f  +  ^  +  f=0,  (6) 

in  which  x',  y',  and  z'  will  be  the  coordinates  of  any  point  on 
any  one  of  the  tangent  lines  to  the  surface  at  P^.  This  equa- 
tion is,  accordingly,  the  equation  of  the  locus  of  the  tangent 
lines  at  the  point  Pi  ;  and  since  this  equation  is  of  the  first 
degree,  the  locus  is  a  plane. 

Transforming  back  to  the  original  axes  by  the  formulas 

x'  =:  X  —  Xi,  y'  =:  y  —  yi,  z'  =  z  —  Zi, 

we  have   j^O^^O  _^  y>  (y  -  yO  ^  zj^z-z^)  ^ 
a"  b"  c'^ 

a^  +  b-.  +  c'^  -  a^'  +  b^  +  c'^  ^^■> 

Therefore,  reducing   by  the    aid  of  (2)  we  have,  as  the 
required  equation  of  the  tangent  plane, 

^+^^r-+t?=k.  [16] 


34.     Normal  Line. 

The  normal  line  at  any  point  of  a  surface  is  the  2)erpen- 
dicular  to  the  tangent  plane  at  that  pioint. 

Now,  the  direction  cosines  of  the  normal  to  the  tangent 
plane  will  be,  by  §  17, 


i^    yi    fi 
a^    b 


z" 


where  R  =  V7?  +  y^  +  ^' 


R     R     R '  >'  a^    '    b^        c" 


334  SOLID    ANALYTIC    GEOMETRY. 

Therefore,  by  [14],  §  23,  the  required  equations  of  the 
normal  line  will  be,  after  the  denominators  have  been  multi- 
plied by  R, 

X  —  Xi  ^  y  —  yi  ^  z  —  Zi 

Xi  Yi  ^  [17] 

a'  b'  c' 


35.     Polar  Plane. 

Let  Pi  (xi,  yi,  Zi)  be  any  point  from  which  secant  lines  are 

x^       y2      z^ 
drawn  to  the  surface  ^  +  f^  +  —  ^  k. 
a^      h^      c 

We  shall  define  the  polar  of  Pi,  ivith  resj^ect  to  this  surface, 
as  the  locus  of  the  line  of  intersectio7i  of  the  tivo  tangent  jjkmes 
to  the  surface  at  the  points  where  it  is  cut  by  any  one  of  the 
secant  lines  from  Pi.  Conversely,  Pi  will  be  called  the  pole  of 
this  locus. 

Let  any  line  from  Pi  meet  the  surface  at  the  points 
P2  (x2?  y2j  Z2)  and  P3  (xs,  y3,  Zg).  The  equations  of  the  line 
P2P3  will  be,  by  [15],  §25, 

X  —  xo  _  y  —  y2  _  z  —  Z2  .^. 

X2  —  X3      y2  —  ys      Z2  —  Z3 ' 


and  since  Pi  is  a  point  of  this  line, 

xi  —  X2  _  yi  —  y2  _  zi  —  Z2 


X2  —  X3      y2  —  ys      Z2  —  Z3 
The  tangent  planes  at  Pg  and  P3  are  respectively 


X3X  I  ysy  ,  Z3Z 

b 


^  +  ^^  +  ^-k  =  0. 


(2) 


(3) 


The  equations  (3)  are  the  equations  of  a  straight  line,  the 
locus  of  which  is  the  polar  of  Pi.     To  find  the  equation  of 


EQUATIONS    OF    SECOND    DEGREE.  335 

this  locus  it  is  necessary  to  eliminate  from  (3)  the  coordi- 
nates, X2,  y2,  Z2,  X3,  ys,  Z3.  This  may  be  done  by  aid  of  (2),  as 
follows  : 

Subtracting  the  second  of  the  equations  (3)  from  the  first, 
we  have 

(X2  —  XQ  X  (y2  -  Ys)  y     ,     (Z-2  -  Z3)  z  ^  ^ 

a^                    b-                     0-               '              ^  ^ 
If  we  multiply  tlie  first  term  of  this  equation  by  — -, 


the  second  term  by  — ^,  and  the  third  term  by  ~ -, 

y2  — ys  Z2  — Z3 

these  multipliers  being  equal  by  virtue  of  (2),  we  have 

(xi  -  xo)  X      (yi  —  ya)  y  ,  (zi  -  zo)  z  ^  ^^ 

32  1^2  (,2 

Finally,  by  adding  to  (5)  the  first  of  the  equations  (3),  we 

as  the  equation  of  the  locus  of  the  intersection  of  the  two  tan- 
gent planes  at  P2  and  P3.  But  this  equation,  which  is  that  of 
a  plane,  is  independent  of  Xj,  y2,  Z2  and  X3,  y3,  Z3,  and  is,  there- 
fore, true  for  every  secant  line  from  Pi. 

■■■'^  +  ^+^  =  ^  [18] 

is  the  polar  of  the  point  Pi  (xi,  yi,  Zi). 


36.     Diametral  Plane. 

The  locus  of  the  middle  2)oints  of  a  system,  of  parallel  chords 
of  the  surface 

a-       b-      c- 


is  a  plane,  called  the  diametral  plane. 


336  SOLID    ANALYTIC    GEOMETRY. 

Let  cos  a,  cos  /?,  COS  y  be  the  direction  cosines  of  each  of 
the  system  of  parallel  chords,  and  let  Pi(xi,  y,,  Zi)  be  the 
middle  point  of  any  one  of  them. 

Transforming  to  P^  as  origin  by  formulas  [7],  §  8,  we  have 
the  equation  of  the  surface  in  the  form 

v'2  v/'2  7'2  9y    y'  '^\/   \l'  ^7    7'  y   ^  ,,   -  7  ^ 

^  +  ^u2  +  -'  +  ^  +  ^+"-^  +  ^  +  T;^  +  ^  =  k-  (1) 

a^       b''       c-        a^  h"  c         a-       b-        c-  ^  ^ 

If,  now,  we  let  x'==:r'  cos  a,  y'  =  r'  cos  (3,  z':=-r'  cos  y, 
(1)  becomes 

/COS^a    i^COS^/S        COS^yX         ^    ^^Xi  COS  a  _|_  yi  COS  ^^  Zj  COS  y 


But  if  cos  a,  COS  /8,  COS  y  are  the  direction  cosines  of  the 
chord  bisected  at  Pi,  the  new  origin,  the  two  values  of  r' 
which  satisfy  this  equation  must  be  numerically  equal  and  of 
opposite  sign,  since  they  are  measured  in  opposite  directions. 
That  such  may  be  the  case,  (2)  must  be  a  pure  quadratic 
equation,  so  that 

Xi  cos  a      yi  cos  j8  1^  Zi  cos  y 

a^  b"  c"  ^  ' 

But  (xi,  yi,  Zi)  may  represent  the  middle  point  of  any  one 
of  the  system  of  chords. 

X  cos  a   ,   y  cos  B   ,    z  cos  V       _  r.^-y 

•■■  ^;r- +  V^  +  ^^  =  0  [19] 

is  the  required  equation  of  the  diametral  plane,  for  we  can 
now  call  this  locus  a  plane,  since  it  is  proved  to  be  a  plane  by 
the  form  of  its  equation. 

As  (0,  0,  0)  satisfies  [19],  it  is  evident  tliat  every  diametral 
plane  passes  through  the  centre  of  the  surface,  and,  conversely, 
it  may  be  proved  that  every  plane  through  the  centre  is  a 
diametral  plane. 


EQUATIONS    OF    SECOND    DEGREE.  337 

37.     Diameters. 

Ani/  straiffht  line  through  the  centre  of  a  surface  is  called 
a  diameter.  It  follows  that  any  two  diauieters  must  neces- 
sarily determine  a  plane  through  the  centre  of  the  surface, 
which  will  be  a  diametral  plane  by  §  36. 

Noiv,  three  diameters  such  that  the  plane  determined  by  two 
of  them  bisects  chords  parallel  to  the  third  are  called  conjxujate 
diameters. 

We  will  now  determine  the  necessary  conditions  that  three 
diameters  of  direction  cosines  cos  ai,  cos  ^i,  cos  yi,  cos  a^, 
cos  /?o,  cos  yo)  ^nd  cos  ag,  cos  ySs,  COS  yg,  respectively,  shall  be 
conjugate  diameters. 

By  [19],  §  36, 

X  cos  ai       y  cos  ^1       z  cos  yi  _ 

^2  -^  b^  -^  c-  ~       '  ^^ 

is  the  diametral  plane  bisecting  chords  parallel  to  the  tirst 
diameter. 

If  these  three  diameters  are  conjugate,  the  second  diameter 
must  lie  in  this  plane,  and  hence  must  be  perpendicuhu*  to 
the  normal  to  the  plane.  Writing  down  the  condition  that 
these  two  lines  shall  be  perpendicular  to  each  other,  and 
reducing,  we  have 

cos  tti  cos  a2    .     cos  /8i  COS  ySo    I    cos  yi  cos  ya „ 

a'^  ^  b^^  ^  ?  -"'  ^"^ 

as  the  necessary  condition  that  the  second  diameter  sluill  be 
in  plane  (1). 

Similarly,  if  the  third  diameter  is  in  plane  (1), 

cos  ag  cos  tti       cos  /?g  COS  /3i   ,  COS  yg  cos  yi 

a-  b""  c^  ^ 

In  like  manner,  if  the  first  and  the  third  diameters  lie  in 
the  plane  bisecting  chords  parallel  to  the  second  diameter,  we 
have 


338  SOLID    ANALYTIC    GEOMETRY. 

COS  ai  COS  az    ,    COS  /8i  COS  ^2    ,    COS  yi  COS  ya r^  ^.^ 

COS  a2  COS  gg        COS  /jg  COS  jgg        COS  ya  COS  yg  _ 

^i  +-        b^  +  ^i  -<^5         (^) 

and,  if  the  first  and  the  second  diameters  lie  in  the  plane 
bisecting  chords  parallel  to  the  third  diameter,  we  have 

cos  as  cos  ai      cos  183  cos  I3i  .  cos  yg  cos  yi 

cos  a2  COS  as        COS  (32  COS  ySg    ,    cos  y2  COS  ys 

^^  ^  b^  ^  ^^  ~  ^^ 

On  reviewing  this  article  we  see  that  there  are  but  three 
distinct  necessary  conditions  for  the  three  diameters  to  be 
conjugate,  i.e., 

cos  ai  cos  a2  ,  cos  (3i   COS  ^2  ,  COS  yi  cos  y2 „ 

I'  +  b^  "^  c'  ~^' 

cos  ttg  cos  Us     ,    COS  ^2  COS  ySg         COS  y2  COS  yg ^ 

^^  +  b^  "^  ^^  ~      ' 

093  ttg  COS  tti        COS  /Ss  COS  ^1        COS  y3  COS  yi 

and  that,  if  these  conditions  are  satisfied,  the  diameters,  taken 
in  pairs,  will  determine  the  three  diametral  planes, 

X  cos  tti       y  cos  /81   .   z  cos  yi 

"^"^        b^~"^       c'       ~    ' 

X  COS  a2   1^  y  COS  (3^  j^z  cos  y2 ^ 


a 
X  cos 


03      y  cos  ^3      z  cos  y3  _ 
.1-      "^       b^      "^       c^      ~    ' 

each  of  which  bisects  all  chords  parallel  to  the  line  of  inter- 
section of  the  other  two. 


EQUATIONS   OF    SECOND   DEGREE.  339 

38.     Auxiliary  Line  of  the  Ellipsoid. 

In  §  109,  Part  I,  we  have  seen  that,  if  x  and  y  are  the  coor- 

X"       v"^ 
dinates  of  any  point  of  the  ellipse  —^  -\-^-=^l^  then  x  =  a  cos  ^ 

3.  D 

and  y  =  b  sin  <^,  tlie  point  being  thus  determined  by  means  of 
its  eccentric  angle. 

Similarly,  if  x,  y,  z  are  the  coordinates  of  any  point  of  the 
ellipsoid 

-+^+-=1 
we  may  place 

x  ^  a  cos  A,     y  =  b  cos  fx,     z  =  c  cos  v, 
where  cos^  A  +  cos*^  /x  +  cos-  v  ^  1, 

this  condition  being  necessary  in  order  that  the  assumed 
values  of  x,  y,  z  may  satisfy  the  equation  of  the  ellipsoid. 
The  quantities  cos  X,  cos  jx,  cos  v  are,  then,  the  direction  co- 
sines of  some  line,  which  we  will  call  the  mixiliarij  line  of 
the  point  (x,  y,  z).  It  is  to  be  noticed  that  the  auxiliary  line 
is  different  for  different  points  of  the  ellipsoid.  The  use  of 
the  auxiliary  line  will  be  found  to  aid  materially  in  the  solu- 
tion of  some  problems,  as  indicated  in  the  two  following  : 

1.  The  auxiliary  lines  corresponding  to  the  extreviities  of 
two  conjugate  diameters  of  an  ellipsoid  a7'e  2i&^2Jendicular  to  each 
other. 

Let  the  extremities  of  tlie  diameters  be  (xj,  yi,  Zj)  and 
(x2,  y2,  Z2),  respectively,  and  the  lengths  of  the  semi-diameters 
be  ri  and  r,,  respectively.     Then  their  direction  cosines  will 

be  — ,  — ,  — ,  and  — ,   — ,  —,  respectively,  and  since  the  diam- 
•"i     Tj     r^  r2     fz    ''2 

eters  are  conjugate  we  have,  by  §  37, 

jLjx^       y^       z^J        g  (1) 

rifa  (    a-^  b^  c-    ^  ^  ^ 

But  if    cos  Xi,  cos  /xj,  cos  vi  and  cos  A2)  cos  fi^,  cos  v^  are 

the  direction  cosines  of  the  corresponding  auxiliary  lines, 


340  SOLID    ANALYTIC    GEOMETRY. 

xi  =  a  COS  Xi,  Yi  =  b  cos  fx-i,  etc.,  so  that  substituting  in  (1) 
and  reducing  we  have 

cos  Xi  cos  A2  +  cos  fjLi  cos  ju.2  +  cos  vi  cos  1/2  =  0,        (2) 
an  equation  proving  the  theorem  stated. 

2.  The  sum  of  the  squares  of  three  conjugate  semi-diameters 
of  an  ellijjsoid  is  constant. 

Let  ai,  bi,  and  Ci  be  conjugate  semi-diameters,  the  extremi- 
ties of  which  are,  respectively,  (x^,  yi,  z^),  (xg,  yz,  z-i),  and 
(xg,  yg,  Z3),  or,  if  we  use  tlie  auxiliary  line,  the  extremities  of 
which  are,  respectively,  (a  cos  Ai,  b  cos  /^i,  c  cos  j/i),  (a  cos  A25 
b  cos  ju.2,  c  cos  V2),  and  (a  cos  A3,  b  cos  jUg,  c  cos  V3). 

Then,  a^^  -\-  hi  -\-  Cj^  :=  a^  cos^  Ai  +  b-  cos^  /xi  -\-  cr  cos^  vi 
4"  a^  cos^  A2  +  b'  cos^  fXi  -\-  c^  cos^  v^ 
+  a^  cos^  A3  -\-  \y  cos^  fi^  +  c^  cos^  V3. 

But,  by  virtue  of  our  first  theorem,  the  three  auxiliary 
lines  are  perpendicular  each  to  the  other  two,  so  that  cos  Ai, 
cos  A2,  cos  A3  may  be  regarded  as  the  direction  cosines  of 
OX  with  respect  to  a  new  set  of  rectangular  axes,  whence 
cos^  Ai  +  cos^  Ao  +  cos^  A3  =  1. 

In  like  manner  cos^  fxi  +  cos^  fxo  -{■  cos^  ju.3  =  1,  and 

COS^  Vi  +  COS^  V2  +  COS^  V3  =  1. 

Therefore  aj^  +  bi^  +  Ci"  =  a^  +  b"  +  c^  and  our  proposition 
is  proved. 

39.     Parallel  Sections. 

Two  conic  sections  are  said  to  be  similar  and  similarly 
■placed,  if  the  coefficients  of  the  quadratic  terms  in  the  equa- 
tions of  the  two  are  the  same  or  differ  only  by  a  constant 
multiplier.  From  this  definition  it  is  easy  to  show  by  aid  of 
chapter  XI,  Part  I,  that  the  axes  of  the  two  conies  are  paral- 
lel and  proportional  in  length,  and  hence  that  the  eccentrici- 
ties are  equal.  In  particular,  any  two  circles  are  similar 
conies,  as  are  also  any  two  parabolas. 


EQUATIONS   OF   SECOND   DEGilEE.  341 

We  shall  now  prove  that  all  parallol  lAane  sections  of  any 
surface  of  the  second  degree  are  similar  conies. 

The  most  general  equation  of  the  second  degree  is 

Ax^  +  By-  +  Cz-  +  2  Fyz  +  2  Gzx  +  2  Hxy  +  2  Lx  +  2  My 
+  2  Nz  +  D  =  0.* 

Now,  if  we  let  z  =  0,  the  equation  of  the  section  made  by 
the  plane  XOY,  is 

Ax-  +  By2  +  2  Hxy  +  2  Lx  +  2  My  +  D  =  0. 

If  we  let  z  =  Zi,  the  equation  of  the  section  made  by  a  plane 
parallel  to  XOY  is 

Ax'^+  By-  +  2  Hxy +  2  Lx  +  2  My  +  Cz^^-f  2  Fyzi+2  HziX 
+  2  Nzi+  D  =  0. 

In  these  two  equations  the  coefficients  of  x^,  xy,  and  y^  are 
the  same.  Hence  sections  parallel  to  the  plane  XOY  are 
similar.  We  may,  however,  take  any  plane  as  the  plane  XOY, 
by  a  proper  transformation  of  coordinates,  without  altering 
the  degree  of  the  equation.  Hence  the  theorem  is  true 
universally. 


40.     Circular  Sections. 

The  problem  of  finding  when  a  plane  section  of  a  central 
quadric  will  be  a  circle  reduces,  by  §  39,  to  the  problem  of 
finding  when  a  section  made  by  a  plane  passing  through  the 
centre  will  be  a  circle.  We  will  solve  this  problem  for  the 
ellipsoid 

K+j'.+^=i.  (1) 

a        b-      c- 

*  It  may  be  shown  that  this  equation  represents  one  of  the  cones, 
the  central  quadrics,  or  the  paraboloids,  mentioned  in  the  text,  except 
in  the  cases  in  which  the  equation  may  be  separated  into  two  linear 
equations,  the  locus  being  then  two  planes. 


342  SOLID    ANALYTIC    GEOMETRY. 

Now,  all  points  which  are  a  constant  distance  r  from  the 
origin  lie  upon  the  sphere 

t+t+'-.=l.  (2) 

r         r  r 

If  such  points  lie  also   upon  the  ellipsoid  (1),  then,  by 
Ex.  1,  chapter  II,  they  lie  upon  the  surface 

which  passes  through  the  origin.  This  equation  (3)  repre- 
sents, in  general,  a  cone,  thus  showing  that  if  points  on  the 
ellipsoid  at  a  constant  distance  from  the  centre  be  joined  to 
the  origin,  the  joining  lines  form  a  cone,  except  in  excep- 
tional cases.  These  exceptional  cases  occur  when  the  equa- 
tion (3)  may  be  factored  into  two  linear  factors,  thus  showing 
the  corresponding  surface  to  consist  of  two  planes.  But  such 
factoring  is  possible  only  when  r  has  such  values  that  one  of 

the  quantities  -^ -j,  r^ ^,  -^ -2  is  zero.     Accordingly,  if 

a  >  b  >  c,  let  us  place  r  =  b  ;  equation  (3)  then  becomes 

b-      aO       ^    U'      b'-' 


7.-rA=^, 


which  factors  into 


W^-i  +  ^i-i-o, 


Vs-^'-^Vi-^==»' 


and 

b^       a'^  ^c^      b^ 

thus  giving  two  planes  through  the  origin,  which  cut  the 
ellipsoid  in  circular  sections.  These  planes  are  determined 
by  the  axis  of  y  and  the  semi-diameters  in  the  plane  XOZ 
which  are  equal  to  b. 

If  Ave  had  made  either  of  the  other  two  coefficients  zero,  the 
equation  (3)  could  not  have  been  factored  into  real  factors, 


EQUATIONS    OF    SECOND   DEGREE.  343 

and  the  reason  is  not  far  to  seek.  If  we  had  placed  r  =  a, 
thus  making  the  coefficient  of  x^  zero,  we  shoukl  have  been 
required  to  draw  in  tlie  phme  YOZ  a  semi-diameter  greater 
than  b,  which  is  the  greatest  semi-diameter  in  that  plane ;  on 
the  other  hand,  if  we  had  made  r  ^  c,  thus  making  the 
coefficient  z^  zero,  we  should  have  been  required  to  draw  in  the 
plane  XOY  a  semi-diameter  less  than  b,  which  is  the  least 
semi-diameter  in  that  plane. 

Treating  the  equation  of  the  unparted  hyperboloid 

i^  _L  y'  _  e!  =  1 

a^  "^  b^      c^ 
in  the  same  way,  we  find  that  the  planes 

cut  the  surface  in  circular  sections,  if  a  >  b,  and  that  conse- 
quently the  planes  pass  through  the  semi-diameter  a. 

If  we  apply  this  same  method  to  the  biparted  hyperboloid 

and  if  b  >  c,  the  only  way  in  which  we  can  reduce  this  equa- 
tion to  the  equation  of  two  planes  is  by  letting  r^  =  —  b^, 
when  the  equation  becomes 

These  planes  will  not  cut  the  hyperboloid  at  all,  but  the 
})lanes  parallel  to  them,  Avhich  do  intersect  the  hyperboloid, 
will  intersect  it  in  circles. 


41.     Umbilic. 

If  the  plane  cutting  the  ellipsoid  in  a  circle  is  moved  away 
from  the  centre,  the  radius  of  the  circle  will  decrease  as  the 


344  SOLID    ANALYTIC    GEOMETRY. 

plane  approaches  a  position  of  tangency  to  the  surface.  The 
point  of  tangency  thus  obtained  as  the  limit  of  a  system  of 
circular  sections  is  called  an  umbilic.  It  is  characterized  by 
the  fact  that  a  plane  indefinitely  near  the  tangent  plane  and 
parallel  to  it  cuts  the  surface  in  a  small  circle. 

X^  y2  jQ, 

Ex.     Find  the  umbilics  of  the  ellipsoid  77^  +  tt  +  t  =i- 

do       y       4 

Equation  (3),  §  40,  is 

<k-\^^  <-}:)- ^'(a-^)-"- 

For  the  planes  of  circular  section  r  =  3,  and  we  get  3x2  _  5^2  —  q,  or 
xV3+zV5  =  0,  -  (1) 

x^/3-z^/5  =  0.  (2) 

If  (xii  yi)  zi)  is  the  umbilic, 

is  the  tangent  plane  at  that  point,  and  by  definition  (3)  must  be  parallel 
to  (1)  or  (2). 


(4) 


If  (3)  is  parallel  to  (1), 

V3_  0__  V5_ 

X]^     yi      £1 

36        9         4 

Solving  equations  (4)  simultaneously  with 

2^  4.  Zi!  J.  £1:  -  1 

36  "^   9        4 
(since  (xi,  yi,  zi)  is  a  point  of  the  surface)  we  find 

xi  =  ±|V6,  yi  =  0,   zi=±|VlO 

as  the  coordinates  of  two  of  the  umbilics. 

If  we  had  used  (2)  instead  of  (1),  we  should  have  found  the  other  two 
umbilics  of  which  the  coordinates  are 


EQUATIONS    OF    SECOND    DEGREE.  345 


EXAMPLES. 

1.  Show  that  any  homogeneous  equation  in  x,  y,  and  z  rep- 
resents a  cone  with  its  vertex  at  the  origin.* 

2.  Find  the  equation  of  a  riglit  circular  cone  of  angular 
opening  60°,  the  origin  being  the  vertex  and  the  axis  of  z  the 
axis. 

3.  Find  the  e(piation  of  a  cone  with  the  origin  as  a  vertex 
and  the  ellipse  2x^  +  3y-  =  6  in  the  plane  z  =  1  as  a  directrix. 

4.  Find  the  equations  of  the  prolate  and  oblate  spheroids 
wliich  may  be  formed  by  tlie  revolution  of  the  ellipse 

2x2  +  .3y-  =  l. 

5.  Show  that  x^  +  y-  +  z-  +  2  Lx  +  2  My  +  2  Nz  +  D  =  0  is 
the  equation  of  a  sphere,  and  iind  its  centre  and  radius. 

6.  Show  that  the  equation  of  the  sphere  having  the  line 
joining  (xj,  yi,  Zi)  and  (x.,,  y.i,  Zo)  as  diameter  may  be  jmt  in 
the  form 

(X  -  xO  (x  -  X,)  +  (y  -  yO  (y  -  y,)  +  (z  -  z.)  (z  -  z,)  =  0. 

7.  Show  that  four  points  are  sufficient  to  determine  a 
sphere. 

8.  Show  that  on  an  unparted  hyperboloid  every  rectilinear 
generator  of  one  system  intersects  every  rectilinear  generator 
of  the  other  system,  by  showing  that  a  plane  may  be  passed 
through  the  two. 

9.  Show  that  on  an  unparted  hyperboloid  no  two  rectilinear 
generators  of  the  same  system  can  intersect. 

10.  Show  that  the  hyperbolic  paraboloid  is  a  ruled  snrface 
covered  by  two  systems  of  rectilinear  generators. 

11.  Show  that  every  tangent  plane  to  a  cone  of  the  second 
order  passes  through  the  vertex  of  the  cone. 

*  A  cone  is  a  surface  generated  by  the  movement  of  a  straight  line 
which  continually  passes  through  a  fixed  point.  It  is  not  necessarily  of 
the  second  degree. 


346  SOLID    ANALYTIC    GEOMETRY. 

12.  Show  that  the  equation  of  the  plane  tangent  to  the 
sphere  x-  +  y^  +  z^  +  2  Lx  +  2  My  +  2  Nz  +  D  =  0  at  the 
point  (xi,  yi,  z^)  is 

XiX  +  yxy  +  ziz+L(x  +  xO+M(y  +  yO  +  N(z  +  zO  +  D  =  0. 

13.  Show  that  the  normal  at  any  point  of  a  sphere  passes 
through  the  centre  of  the  sphere. 

14.  Find  the  equation  of  the  tangent  plane  to  the  parabo- 

X^  y2 

loid  —  ±  f^  =  4pz,  the  point  of  contact  being  (xi,  yi,  Zj). 
a        D" 

15.  Prove  that  the  plane  x  cos  a  +  y  cos  /?  +  z  cos  y  =  p  is 

9  9  '' 

X        y        z" 
tangent  to  the  ellipsoid  — ,  +  f^  -| — ^  =  1  if 
a"^      b''      c" 


p  =  Va^  cos^  a  +  b^  cos^  /3  +  c^  cos^  y. 

16.  Prove  that  the  plane   x  cos  a  +  y  cos  (i-\-  z  cos  y  =  pi 

9  9 

X        y 
IS  tangent  to  the  paraboloid  —,+  7-2^  4pz,  if 

3."         D" 

a^  cos^  a  +  b^  cos^  B 
Pi  = — p. 

^  cos  y  ^ 

17.  Show  that  the  length  of  the  tangent  drawn  from  any 
point  (xj,  yi,  Zj)  to  the  sphere 

x^  +  y-  +  z-  +  2  Lx  +  2  My  +  2  Nz  +  D  =^  0 

is  Vxr  +  yr  +  z,'  +  2  Lxi  +  2  Myi  +  2  Nzi  +  D. 

18.  Show  that  the  locus  of  the  points  from  which  equal 
tangents  can  be  drawn  to  the  spheres 

x2+y2  +  z-  +  2  L'x  +  2  M'y  +  2  N'z  +  D'  =  0 
and       x-  +  y-  +  z-  +  2  L"x  +  2  M"y  +  2  N"z  +  D"  =  0 
is  the  plane 
2  (U-  L")  x  +  2  (M' -  M")  y  +  2  (N'  -  N")  z  +  (D'  -  D")  =  0, 

which  is  called  the  radical  plane  of  the  two  spheres. 

19.  Show  that  the  radical  planes  of  three  spheres  meet  in 
a  single  line,  called  the  radical  line  of  the  spheres. 


EQUATIONS    OF    SECOND    DEGREE.  347 

20.  Show  that  three  mutually  perpendicular  tangent  planes 

x'^       y"      z^ 
to  the  ellipsoid  —„  -\-  ,  .-,  4-  —  =  1  intersect  on  the  sphere 
a^      b"      c^ 

X-  +  y^'  +  z-  =  a-  +  b-  +  c^. 

21.  Show  tliat  three  mutually  perpendicular  tangent  planes 

x^       y2 
to  the  paraboloid  —  +  ,  „  =  4pz  intersect  on  the  plane 
a-      b- 

z-F(a-+b^)  p  =  0. 

22.  Show  that  the  two  rectilinear  generators  through  any 
point  of  an  unparted  hyperboloid  lie  in  the  tangent  plane  at 
that  point. 

23.  Show  that  the  two  rectilinear  generators  through  any 
point  of  an  hyperbolic  paraboloid  lie  in  the  tangent  plane  at 
that  point. 

24.  Find  the  equation  of  the  polar  plane  of  (xi,  y,,  z,)  with 
respect  to  the  sphere  x^  +  y^  +  z^  ^  a^ 

25.  Show  that  the  polar  plane  with  respect  to  a  s})here  is 
perpendicular  to  the  line  joining  the  pole  to  the  centre  of  the 
si)here. 

26.  Show  that  the  product  of  the  distance  of  the  pole  and 
its  polar  plane  from  the  centre  of  the  sphere  is  equal  to  the 
square  of  the  radius  of  the  sphere. 

27.  Find  the  pole  of  the  plane  Ax  +  By  +  Cz  +  D  =  0  with 
respect  to  the  sphere  x^  +  y-  +  z"  =  a'-. 

28.  Find  the  pole  of  the  plane  Ax  +  By  +  Cz  +  D  =  0  witli 

9  '2  2 

x        y        z 
respect  to  the  ellipsoid  —,-\-t7,-\ — r,  =  1- 
a-      b-      c- 

29.  Show  that  any  secant  from  Pi  to  an  ellipsoid  is  divided 
harmonically  by  Pi,  the  })olar  plane  of  Pi,  and  the  surface  of 
the  ellipsoid. 

30.  Show  that  the  polar  of  any  point  of  a  plane  with  re- 
spect to  a  central  quadric  passes  through  the  pole  of  the 
plane. 


348  SOLID    ANALYTIC    GEOMETRY. 

31.  The  intersection  of  the  polar  planes  of  A  and  B  is  a 
line  CD,  which  is  called  the  polar  line  of  AB.  Prove  that 
the  polar  planes  of  all  points  on  AB  pass  through  CD,  and 
that  the  polar  planes  of  all  points  on  CD  pass  through  AB.* 

32.  ABCD  is  any  tetrahedron  and  A'B'C'D'  is  a  second 
tetrahedron  such  that  its  faces  B'C'D',  C'D'A',  etc.,  are  the 
respective  polars  of  A,  B,  etc.  Show  that  A'  is  the  pole  of 
BCD,  B'  the  pole  of  CD  A,  etc.,  or  that  the  tetrahedrons  are 
conjugate.! 

33.  Show  that  the  edges  AB  and  CD'  in  example  32  are 
reciprocal  polar  lines. 

34.  Find  the  equation  of  the  diametral  plane  to  the  sphere 
x2  _[-  y2  _|_  ^2  =:  ^2^  g^ud  sliow  that  it  })asses  through  the  centre 
of  the  sjjhere  and  is  perpendicular  to  the  chords  which  it 
bisects. 

35.  Find  the  equation  of  the  diametral  jjlane  of  the  parabo- 

9  9 

x"       y 
loid  ~dz~  =  4pz,  bisecting  the  cliords  the  direction  cosines 
a''      b- 

of  which  are  cos  a,  cos  /3,  cos  y. 

36.  Find  the  direction  cosines  of  the  chords  of  the  un parted 

X^  y2  2^ 

hyperboloid  —  +  f:, ^  =  1,  which  are  bisected  by  the  plane 

Ax  +  By  +  Cz  =  0. 

37.  Show  that  all  planes  parallel  to  the  axis  OZ  of  a  parab- 
oloid are  diametral  planes. 

38.  Prove  that  every  plane  passed  through  the  centre  of 
a  central  quadric  is  a  diametral  plane  and  determine  the 
direction  cosines  of  the  chords  bisected. 

39.  Show  that  the  tangent  plane  of  a  central  quadric  is 
parallel  to  the  diametral  plane  bisecting  chords  parallel  to  the 
diameter  drawn  to  the  point  of  contact. 

*  Two  such  lines  are  called  reciprocal  polar  lines. 
t  A  tetrahedron  is  self-conjugate,  if  each  vertex  is  the  pole  of  the  oppo- 
site face. 


EQUATIOXS    OF    SECOND    DEGREE.  349 

40.  Show  that  a  tangent  plane  to  the  paraboloid 

9  O 

^  _i_  y     -( 

is  parallel  to  all  the  chords  which  are  bisected  by  the 
diametral  planes,  the  line  of  intersection  of  which  passes 
through  the  point  of  tangency. 

41.  Show  that  of  three  conjugate  diameters  of  a  sphere 
each  is  perpendicular  to  the  other  two. 

42.  Show  that  the  tangent  planes  at  the  extremities  of 
three  mutually  perpendiciilar  diameters  of  a  sphere  intersect 
on  a  concentric  sphere,  the  radius  of  which  is  VS  times  the 
radius  of  the  first  sphere. 

43.  Show  that  the  tangent  planes  at  the  extremities  of  con- 
jugate diameters  of  the  ellipsoid  intersect  on  the  surface 

^  +  ^  +  ^  =  3. 
a-       b-      c" 

44.  If  tangent  planes  are  drawn  at  both  ends  of  each  of 
three  conjugate  diameters  of  an  ellipsoid,  show  that  they  are 
the  faces  of  a  circumscribing  parallelopiped. 

45.  Show  that  the  sum  of  the  reciprocals  of  the  squares  of 
three  semi-diameters  which  are  mutually  at  right  angles  is 

constant  and  equal  to  —  -I-  —  H — 5 
a-       b-      c 

46.  Show  that  the  sum  of  the  squares  of  the  projections 
of  three  conjugate  semi-diameters  of  an  ellipsoid  on  a  given 
straight  line  having  cos  a,  cos  (3,  cos  y  as  direction  cosines  is 
constant  and  equal  to  a"  cos^  a  +  b^  cos^  /?  +  c^  cos-  y. 

47.  Show  that  the  sum  of  the  squares  of  the  projections  of 
three  conjugate  semi-diameters  on  any  plane  is  constant. 

48.  Show  that  if  two  diametral  planes, 

X  cos  ai   ,   y  cos  (3i 

7, \-  - — 7J> 2p  cos  yi  =  0 

a-  b 

-                   X  cos  a.2   ,   y  cos  B2      0  n 

and  —^^ \-  —^  -  -P  t'os  y.  =  0, 


350  SOLID    ANALYTIC    GEOMETRY. 

of  the  parabolid  -i,-\-  h  =  4pz,  are  such  that 
a"      b" 

COS  tti  cos  ag    ,    cos  /Si  COS  (32 ^ 

a^  +  b^  ~    ' 

then  each  phine  bisects  all  chords  parallel  to  the  other. 
49.  Find  the  umbilics  of  the  hyperboloid 

x^      y^       z^  

4  ~  9  ~3G~    ■ 


FOEMULAS. 

(Rectangular  coordiuates  are  used  unless  otherwise  stated.) 

PART   I. 
PLANE  ANALYTIC  GEOMETRY. 
Distance  between  two  points  : 


d=  V(xi-x,)^  +  (yi-y2)^.  [1],  §3. 

Slope  of  line  through  two  points  : 


^  =  yL_y^.  [2],  §3. 

Xi  X2 


Point  of  internal  division  : 


I1X.2+  I2X1      _  Iiy2  +  l-iYi 
I.  +  I2    '  ^         I1  +  I2    ' 


[3],  §4. 


Point  of  bisection  : 


x  =  li±^,y  =  >4X..  [4],  §4. 


Point  of  external  division  : 

I1X2—  I2X1       _  Iiy2—  l2yi 


'^^V=V''^^2-        ™'§*- 


Area  of  triangle  : 


Area  PiPgPs^  i{(xiy2  —  Xoy,)  +  (xoya  —  x^y^) 

+  (x:,yi-x,y3)}.  [6],  §5. 


352  PLANE   ANALYTIC    GEOMETRY. 

Relations  between  rectangular  and  polar  coordinates  : 

X  =  r  cos  ^,  y  =  r  sin  6.  [7],  §  10. 

r  =  Vx^  +  f,  0  =  tan- 1  ^ •  [8],  §  10. 

Equations  of  the  straight  line  : 

y  =  mx-|-b.  [9],  §21. 

^  +  ^  =  1.  [10],  §22. 

X  cos  a  +  y  sin  a=  p.  [11],  §23. 

r  cos  (^  — a)  =  p.  [12],  §31. 

Line  through  a  point  with  a  given  slope  : 

y-y,=  m(x-xi).  [13],  §33. 

Line  through  two  points  : 


yj-yi  _  X  —  Xi 
yi  — y2     xj  — X2 


[14],  §34. 


Angle  between  two  lines  : 


tan  /?  =  ^^ ^  •  [15],  §  35. 

Parallelism  :                          trii  ^  nria.  [16]?  §  35. 

Perpendicularity  :  rrii  = [17],  §  35. 

Line  through  a  point  perpendicular  to  a  given  line  : 

y-yi  =  --(x-xO.  [18],  §36. 

Translation  of  origin  : 

x  =  Xo+x',  y  =  yo  +  y'.  [19],  §43. 

Rotation  of  rectangular  axes  through  an  angle  6  : 

X  =  X' cos  «-/ sine  ^ 

y  =  x' sm  ^-|- y' COS  ^. 


FORMULAS.  353 

Transformation  from  rectangular  to  oblique  axes   with  the 
same  origin  : 

x  =  x'cos^  +  y'cos^', 

y  =  x'  sin  ^  +  y'  sin  6'.  ^     ^'  ^ 

Transformation  from  oblique  to  oblique  axes  with  the  same 
origin  : 

_  x'  sin  (co  —  ^)  +  y'  sin  (o)  —  0') 
sin  CO  ' 

,.,,,.„  [22],  §46. 

X   sm  y  +  y  sin  6 

y—    __/ . 

^  sin  w 

Equations  of  the  circle  : 

(x-d)-+('y-e)2  =  rl  [23],  §  49. 

x^'  +  y^'=r.  [24],  §49. 

x2  +  y2  +  2  Gx  +  2  Fy  +  C  =  0.         [26],  §  50. 

Tangent  to  circle  : 

xix  +  yiy  +  G  (x  +  xO  +  F  (y  +  y,)  +  C  =  0.     [26],  §  54. 
Polar  with  respect  to  the  circle  : 

xix  +  y.y  +  G  (x  +  xO  +  F  (y  +  y,)  +  C  =  0.     [27],  §  59. 
Diameter  of  circle  : 

y-e  =  -^(x-d).  [28],§61. 

Polar  equation  of  the  circle  : 

r  +  rf  —  2rri  cos  {6  -  6,)  =  a'^.         [29],  §  65. 
Equation  of  conic  section  : 

(x-2p)2+y^-eV.  [30],  §  66. 

Simplest  equation  of  the  parabola  : 

r  =  4px.  [31],  §67. 


354  PLANE   ANALYTIC    GEOMETRY. 

Simplest  equation  of  the  ellipse  : 

^,  +  ^■=1.  [32],  §68. 

Simplest  equation  of  the  hyperbola  : 

|-^,  =  1.  [33],  §69. 

Tangent  to  conic  section  : 

Axix  +  Byiy  +  G  (x  +  x^)  +  F  (y  +  yO  +  C  =  0.    [34],  §  76. 
Polar  with  respect  to  the  conic  section  : 

Axix  +  Byiy  +  G  (x  +  xO  +  F  (y  +  Yi)  +  C  =  0.    [35],  §  78. 
Diameter  of  parabola : 

y  =  ^-  '   [36],  §87. 

Parabola  referred  to  diameter  and  tangent  as  axes  : 

y2  =  4p'x.  [37],  §90. 

Polar  equation  of  the  parabola  : 

2p 

r  = ' • 

1  —  cos  6 
Diameter  of  ellipse  : 


[38],  §  91. 


y  =  -  ^  x.  [39],  §  100. 


Condition  for  conjugate  diameters  of  ellipse 


n^^rr^,  =  --,'  [40],  §  101. 

a 


Equations  of  the  ellipse  in  terms  of  the  eccentric  angle  : 

=  a  cos  <^,  y  ==  b  sin  <^,  [41],  §  109. 


X 


FORMULAS.  355 

Ellipse  referred  to  conjugate  diameters  as  axes  : 

^^  +  ^.  =  1-  [42],  §111. 

Polar  equation  of  the  ellipse  : 

''=^ ^— ^^-  [43],  §112. 

1  —  6'  COS''  6  L       J'    o 

Diameter  of  hyperbola  : 

y  =  ^    X.  [44],  §119. 

Condition  for  conjugate  diameters  of  hyperbola  : 

mim2  =  -2.  [45],  §120. 

Hyperbola  referred  to  conjugate  diameters  as  axes  : 

^,1-^,  =  1.  [46],  §121. 

Hyperbola  referred  to  the  asymptotes  as  axes  : 

a-  +  b' 
xy  =      ^      .  [47],  §  127. 

Polar  equation  of  the  hyperbola  : 

''  =  -2 ?^ T  •  [48],  §  128. 

6''  COS^  6—1  L       J^   o 

Keduction  of  the  general  equation  of  the  second  degree,  when 
AB-  H'<0: 

To  remove  the  terms  of  the  first  degree,  transfer  the 
origin  to  (xq,  yo),  where 

Axo+Hy„+G-0, 
Hxo+  By„+  F=0. 


356 


PLANE   ANALYTIC    GEOMETRY. 


To  remove  the  xy  term,  rotate  the  axes  through  an 
angle  6,  where 

2  H 

«  =  *»--' a3^; 

the  coefficients  A',  B',  C  of  the  resulting  equation 

A'x"^+  B'y"-  +  C'  =  0 

will  be  determined  from  the  equations 

C  =  Ax,/  +  2  Hx„yo  +  By,2  +  2  Gx„  +  2  Fy.,  +  C, 

A'+  B'=  A+  B, 

A'B'=  AB-  H-, 

where  A'—  B'  shall  have  the  same  sign  as  H. 

Reduction  of  the  general  equation  of  the  second  degree,  when 

AB-  H2=0. 

To  remove  the  xy  term,  rotate  the  axes  through  such 
an  angle  that  the  equation  of  OX'  shall  be 

VAx+  VBy  =  0, 
,_  VBx  —  VAy 

'  ~    Va+~b~' 

,_  VAx+VBy 

'  ~    Va+b   ■ 


and 


The  reduction  may  now  be  finished  by  the  method  of 
completing  the  squares. 


FOEMULAS.  357 

PART   II. 
SOLID  ANALYTIC  GEOMETRY. 

Distance  between  two  points  : 


d-V(xi-x,)^+(y,-y,)^+(zi-Z2)^.      [1],  §3. 
Point  of  division  : 

I1X0  +  I2X1  l,yo+loyi  liz„+l,z, 

Direction  cosines  : 

COS^a+COS^^  +  COS'^y^l.  [3],  §  5. 

Angle  between  tw^o  lines  : 

COS  6  =  COS  ttj  cos  a^  +  COS  ^j  COS  (3^  +  COS  yi  COS  y2-  [4],  §  7. 

Perpendicularity  : 

COS  ai  cos  a2  +  COS  /3i  COS  ^2  +  COS  yj  COS  y^  =  0.       [5],  §  7. 

Parallelism : 

ai  =  a2,   /3i  =  (32,   71  =  72-  [6],  §  7. 

Translation  of  origin  : 

X  =  Xo  +  x',    y  =  yo  +  y',    z  =  Zq  +  z'-  [7],  §  8. 

Rotation  of  rectangular  axes  : 

X  ^  x'  cos  tti  +  y'  cos  a2  +  z'  cos  as, 

y  =  x'  cos  /?i  +  y'  cos  /?2  +  z'  cos  /Sj,         [8],  §  9. 

Z  =  x'  cos  yi  +  y'  COS  yo  +  z'  COS  ys- 

Relations  between  rectangular  and  polar  coordinates  : 

z  =  r  cos  (f>, 

X  =  r  sin  <^  cos  6,  [9],  §  11. 

y  =  r  sin  <^  sin  6. 


358  SOLID   ANALYTIC    GEOMETRY. 

Equations  of  the  plane  : 

Ax+By  +  Cz+D  =  0.  [10],  §16. 

X  cos  a  + y  cos /5  + z  cos  y  =  p.     [11],  §  17. 

-  +  ^  +  -  =  1-  [12],§18. 

a       b       c  L     j^  o 

Equations  of  the  straight  line  : 


Ax  +  By  +  Cz  +  D  =  0, 
A'x+  B'y  +  C'z+  D'  =  0. 

yi 


[13],  §  22. 
[14],  §23. 
[15],  §25. 

§27. 
§28. 
§28. 


X"       y        z" 
Biparted  Hyperboloid  :      ^  ~  r^ 2  —  1-  §  28. 

2  2 

Elliptic  Paraboloid  :     ^  +  f-,  =  4pz.  §31. 

a^       h- 

x^       f 
Hyperbolic  Paraboloid  :     ~i~h^ 4pz.  §  31. 

a        b 


cos 

a           cos  /?           cos  y  ' 

x  — 

Xi  _  y  —  yi_  z  — Zi 

Xi- 

-  X2       yi  —  y2       Zi  —  Z2 

Equations  of  the  second  degree  : 

Cone  : 

9                     9                     0 

Ellipsoid 

TTT^T^Ol.fr:>^^ 

TTttv 

9                      9                       9 

a^  ^  b^  ^  c^ 

Tangent  plane  : 


^  +  f+7»=''-  [16],  §33. 


FORMULAS.  369 

Normal  line  : 

_i  Zi  Zi 

d?  b^  c^ 

Polar  plane  : 

f  +  ^^  +  f  =  l'.  [18],  §35. 


Diametral  plane  : 

[19],  §36. 


X  cos  a  ,   y  cos  y8      z  cos  y 


PART   I. 
CHAPTER   I. 

1.  2V4T,  |.      2.    V73,   -|.      3.    2V5,  -|;    Vri,  -|;    V26,  -5. 

4.  Vl3,  I;   V4T,   f;  7  ^^,  1.  7.    2  (a  +  b  +  Va^Tb^). 

10.  (V-,  V).  11.    (-  -¥,  -  i)-  12.    (0,  I). 

13.  (f ,   V),  (V,  ¥)•  14-    (24,  18).  15.   (-  5,  21). 

16.  (13,27).  17.   (-11,   -16).  18.   (7,27). 

19.  4,  V37,  Vei.  20.    V37,  6.  21.    Vl9,  ^V3. 

22.  1.  23.    31.  24.    18.  25.5:3.  26.  11^,  2J. 

27.    3V5  +  2  V3  +V26  -  5V3  +  V53  -  14  V3. 

30.  Vl9,  3V3,   V9T;   fV3. 

31.  V1O-3V34- V34-I5V3 +  ^^;    ^(18-5V3). 
35.    (-f,  1)^  36.    (±|V2,  =Fl^)- 

37.    (±|fV34,   ±i|V34).  38.    (f,  3f). 


CHAPTER   II. 

31. 

y  =  4x.                         32.    6x  -  3y  +  14  =  0,  6x  -  y  +  10  =  0. 

33. 

3x  —  7y  +  2  =  0.        34.    5x  +  4y  +  24  =  0,  5x  +  4y  -  26  =  0. 

35. 

x2  +  y2  +  2x  +  4y-20  =  0.                  36.    y2-8x  +  16  =  0. 

37. 

r  =  tan  6.               38.    r  =  a^.              39.    x2  -  3y2  -  8y  +  16  =  0. 

40. 

5x2  +  5y2  _  50x  -  46y  +  22  =  0. 

41. 

5x2  _  4y2  +  24x  +  16y  -  52  =  0.                    42.   (-  f ,  \). 

43. 

(23,  -  16).                  44.    (-  1,  0).                 45.    (0,  0). 

46.    (tV,  H),  (-3,  -3)._  47.    (3,  -3±2V6).         48.   (-3^5,4). 

51.  .(5±W5,  20t4V5-^  52.    (- ffVio,  /,V^). 

53.    (-H,  -3f).  56.    (±5,  ±4),  (±4,   ±6). 

57.    (-  2,  1),   (-  H,  -  If).  58.    (±  5,  0).  59.    ±  7  VlO. 


362  PLANE  ANALYTIC    GEOMETRY. 

60.    ±7.  61.    f.  62.    ±iV6.  63.    ±fV2. 

66.    (IxV,  Hi)-  67.    (±  15,  zf  5).    _  68.    (12^,  0). 

69.    (2i,  21),   (-11,  If).  70.    (±V5,   ±2V5).  71.    (4,  3±V2T). 


a 

13.  h 

14.  2, 
15  i-O 

16.  0, 

17.  -3, 

18.  00, 

19.  y  =  X  +  3.  20.    2y  =  (-jQ-  V2)x  -  4.  21.    x  +  y  =  0. 
22.  2x  + (V3  + l)y  =4V2.               23.    x  Vg  +  2y  -  10  V2  =  0. 

25.  a  =  -  i,  b  =  i,  p  = 


CHAPTER  IIL 

b 

P 

m 

cos  a 

sin  a 

-1, 

1 

Vio' 

3, 

3 

Vio' 

1 

Vio 

-5, 

10 

V29' 

2» 

5 
V29' 

2 
V29 

2> 

0, 

2, 
0, 

3 

-2, 

h 

|V5, 

4. 

W5. 

CO,. 

3, 

CO, 

-1, 

0. 

h 

h 

0, 

0, 

1. 

2V13  +  6V3' 


-  3  .  1 

L,  sin  a 


2V13  +  6V3  V13  +  6V3 

28.    -%=  •  29.    ^  VlO.  30.    Vl3.  31.    ^\  VlO. 

V37 

32.  3x  +  4y  -  20  =  0,  3x  +  4y  +  10  =  0. 

33.  2x +  y +  2±2V5  =  0.  34.    2x  +  y  -  3  =:  0,   x  -  2y  +  4  =  0. 
35.    4x  -  3y  -  3  =  0.               36.    x  -  y  -  3  =  0  ;  (3,  0),  (0,  -  3). 

q  17  20 

37.     2x  +  5y  +  17  =  0;   -^,  -iL,  ^-  38.    f. 

V29     V29     V29 

39.  3x  -  2y  -  4  =  0.  40.    llx  -  3y  +  18  =  0. 

41.  2x  —  y  —  7  =  0.  42.    X  +  1  =  8.  43.    x  -  y  +  2  =  0. 

44.  tan~^  if     45.  tan~^  |.     46.  5x  +  3y  +  4  =  0. 

47.  X  —  2y  =  0.  48.  30x  +  24y  —  35  =  0. 

49.  4x  -  14y  +  5  =  0.     50.  5x  -  27y  +  5  =  0,  15x  -  23y  +  15  =  0. 

51.  21x  +  y  +  101  =  0,  19x  -  9y  +  131  =  0. 

52.  3x  -  28y  -  37  =  0,  27x  -  8y  -  89  =  0. 

53.  7x  -  24y  +  17  =  0,  x  -  1  =  0. 

54.  8x  -  7y  -  59  =  0,  8x  +  7y  +  11  =  0.        55.  x  +  y  =  0. 
56.  29x  +  32y  +  15  =  0.        57.  4x  -  4y  -  9  =  0. 


ANSWERS.  363 

58.  161x  -  4(;y  —  88  =  0.  59.    4x  +  3y  -  1 7  =  0. 

60.  x-y-l  =  0.  61.    2x-5y-41=0. 

62.  X  +  y  -  9  =  0.  63.    8x  -  y  -  17  =  0. 

64.  25x  -  3y  +  3  =  0,  31x  +  78y  -  190  =  0,  37x  -  122y  +  10  =  0. 

65.  3.  66.    99x-27y +  61  =  0,  21x +  77y- 121  =0. 

67.  (1  ±  V3)x  +  (2  ±  V2)y  +  5=0. 

68.  (15  ±  4  ViO)x  +  (5  qi  3  Vl7))y  =  0.  71.    VSS,  -^L  • 

V58 

72.    (2,  0),  (12,  0),  (5,  7) ;  tan"^  |,  45°,  tan"'  §.        73.    f  Vil,  |,  §  VtI. 

74.    2x  -  y  —  8  =  0,  llx  -  lOy  -  17  =  0,  x  —  5y  -  22  =  0. 

9         45         45 
75     '  — 1  — — • 

V5     V22I      V2G 

76.  2x  +  4y  -  13  =  0,  20x  +  22y  -  51  =  0,  5x  +  y  +  7  =  0. 

77.  (—  2^5,,  4^8).  78.     V29  +  2  Vn  +  5,  13. 

79.    llx +  6y-87  =  0,  x-.3y-2  =  0,  X -7  =0.  80.    (7,  |). 

81.  2x  +  5y  -  8  =  0,  X  -  4y  -  1  =  0,  4x  -  .3y  -  10  =  0. 

82.  8x  +  2y  -  55  =  0,  3x  +  4y  —  32  =  0,  5x  — 2y  -  30  =  0. 

83.  e  =  c.         86.  ((V6-V2)a,  I).        87.  (^^  a,  |  W^  a,  ^J^V 

88.    12x  -  7y  +  8  =  0,  Vl93.  89.    (-  14^,  9^). 

91.    (-  2  j-\,  2^1).  102.    X  +  y  =  0,  2x  +  y  -  1  =  0. 

103.    3x  -  y  +  3  =  0,  x  +  3y  +  1  =  0.  108.    tan~^ 2  VH-^  -  AB  . 

'  '  A+  B 


CHAPTER   IV. 

1.  3x  +  4y  =  0.  2.  X-  +  y2  =  10.                3.  y2  =  x''  -  x. 

4.  y2  =  8x.  5.  xy  =  —  8  or  xy  =  8.    7.  3xy  =  1. 

8.  2x2  +  2y2  =  25.  9.  x2  =  4y.                     10.  4x2  —  6y2  =  3. 

11.  12x2  +  16y2  =  51.  12.  3x2  +  4y  =  0.            13.  xy  =  0. 

14.  3x3  +  ^y'^  -  ^-  ^6-  ■■  ^"^  (^  ~  '^)  ~  p- 

17.    r2  sin  2  61  =  14.  18.    r-^  =  -, ,   ,^"^"  . -^  ■ 

a —  (a —  b-)  cos-  6 

2  a  sin2^ 

19.    r  =  8a  (cos  ^  + sin  6).      20.    r  = -— •  21.    r  =  a  tan  ^. 

^  cos  6 

22.    x  +  y=10^^.  23.    x  =4V3.  24.    x- +  y'- -  ay  =  0. 

25.    (x2  +  y2  +  ax)2  =  a2(x2  +  y-).  26.    (x2  +  y2)3  =  4  a2x-y2. 

27.    (x2  +  y2)2  =  a2 (x2  -  y2).  28.    (x-'  +  y2)3  =  a2  (x-'  -  y2  +  2xy)2. 


22. 


364  PLANE    ANALYTIC    GEOMETRY. 


CHAPTER   V. 

5.  x2  +  y2  +  2x  —  12y  +  32  =  0. 

6.  x2  +  y2  -  4x  +  6y  +  8  =  0  ;    x2  +  y^  -  Ox  +  2y  +  5  =  0. 

7.  x2  +  y2  +  3x  —  y  -  10  =  0.  8.    x^  +  y^  =  a^  +  b^. 
9.  x2  +  y2  +  4x  —  2y  =  0.                    10.    x'^  +  y^  ±  2ax  =  0. 

11.  x2  -f-  y2  ±  2ax  ±  2ay  +  a^  =  0.      12.    25x-'  +  25y2  =^  49. 

13.  5x2  +  5y2  +  lox  -  20y  -|-  24  =  0. 

14.  13x2  +  i.3y2  _  7yx  +  52y  +  151  =  0. 

15.  x2  +  y2  +  4x  +  4y  +  4  =  0  ;    9x2  +  9y2  +  12x  -  12y  +  4  =  0. 

16.  (3,   1);    Vn.  17.    (-2,  5);   2.  18.    (2,  -3);   2V3- 
19.  (1,   -1);    iVl41-        20.    (I,   -I);    Vt.  21.    (i,   - 1) ;   0. 

/_G^   _FN      VG'^+F2-CA.  23_    6^_5    +3^0. 

\      A  A  /  A 

25.  x2  +  y2  -  19x  -  17  y  =  0.  26.    3x2  +  ,3y2  _  5x  +  y  _  12  =  0. 

27.  3x2  +  3y2  +  2x  +  2y  —  10  =  0.  28.    x2  +  y2  =  a2  +  b2. 

29.  19x2  +  10y2  —  65x  —  55y  —  50  =  0.        30.    x2  +  y2  -  8x  -  9  =  0. 

31.  x2  +  y2  -  3x  +  4y  -  31  =  0.        32.    2x2  +  2y2  +  2x  -  y  -  15  =  0. 

33.  x2  +  y2  — 3x  —  lly  —  40  =  0. 

34.  x2  +  y2  -  8x  -  20y  +  31  =  0 ;    x2  +  y2  +  lOx  +  lOy  -  35  =  0. 

35.  5x2  4.  j-jyi  _  22x  —  6y  -  39  =  0  ;    5  x2  +  5y2  -  8x  -  4y  —  61  =  0. 

36.  x2  +  y2  +  2x  -  14y  +  41  =  0.  37.    x  +  y  =  0. 
38.  2  (a  -  a')  X  +  2  (b  -  b')  y  =  a2  -  a'2  +  b2  -  b'2. 

40.  x2  +  y2  +  4x  —  6y  —  36  =  0  ;    x2  +  y2  -  80x  +  50y  +  1000  =  0. 

41.  3x2  +  3y2  +  5x  _  5y  _  20  =  0  ; 

40x2  +  40y2  _  400x  +  520y  +  2429  =  0. 

42.  3x2  +  3y2  +  lly  -  17  =  0.      43.    3^2  +  3y2  +  4x  —  4y  —  19  :=  0. 
44.  x2  + y2-2x +  4y  —  20  =  0.   45.    yV3  =  x±4V3. 

46.  y  =  xV3  +  2V3  +  7;   y=xV3+2V3-l. 

47.  6x  —  8y  +  49  =  0  ;   6x  —  8y  —  71  =  0.  48.    x  -  2y  ±  5  =  0. 
49.  3x  +  2y  +  1  ±  3  Vl3  =  0.  50.    3x  -  4y  ±  20  =  0. 
51.  0,  —  f                          52.    y -e=  m(x  —  cl)±  rVl  +m2. 

53.  y  +  F  =  m  (x  +  G)  ±  V(G2  +  F2-  C)  (1  +  m2). 

54.  2x  -  3y  —  13  =  0  ;    3x  +  2y  =  0.  55.    y  +  3  =  0  ;    x  -  2  =  0. 

56.  3x  -  4y  +  50  =  0  ;   4x  +  3y  +  25  =  0. 

57.  X  —  2y  =  0;   2x  —  y  —  3  =  0.        58.    4x  +  2y  Vs  -  12  -  3V5  =  0. 
59.  2x  -  y  +  2  =  0.  60.    Vl7.  61.    V6  ;   0. 

62.  lOx  -  7y  +  4  =  0.  66.    i^^,  i).  68.    ^  • 


ANSWERS.  365 

70.  I  Vs.  71.    X  +  3  y  -  10  =  0  ;    2  V39.  72.    x  +  2y  +  3  =  0. 

73.  12x  +  12y  V3  -  3  V3  +  2  =  0.  74.    x  -  y  +  6  =  0. 

75.  6x  +  4y  -  1  =  0.  76.    5x  +  4y  -  1  =  0 ;    *. 

77.  21x  -  12y  +  1  =  0.        78.    14x  -  20y  +  5  =  0  ;    lOx  +  7y  +  1  =  0. 

79.  (^,  2^5)-  80.    (If,  -If). 

81.    13x2  +  i3y2  +  24x  —  Gy  —  184  =  0.  89.    A  concentric  circle. 

a2 
92.    x-  +  y'"  =  X  ■  ^"^^   ^  circle  through  0. 

94.    A  straight  line  perpendicular  to  the  line  joining  the  point  to  the 

centre  of  the  circle. 
98.    A  circle. 

CHAPTER   VI. 
1.    5x2  +  9y2  _  72x  +  144  =  0.  2.    3x2  _  y2  _  lOx  -  25  =  0. 

3.    x2  -  4y  +  4  =:  0.  4.    8x2  +  9y2  _  igx  -  .3Gy  +  45  =  0. 

5.  25x2  _  iiy2  _  i50x  +  272y  +  589  =  0. 

6.  y2  —  4x  -  Oy  —  3  =  0.  7.    x2  =  4y.  8.    3x2  +  gy  =  q. 
9.    y2  +  12x  =  0.            10.    y2  —  IGx  =  0.             11.    x2  +  8y  -  10  =  0. 

12.  y2  +  20x  -  (iy  -  51  =  0.  13.    x2  —  2x  -  4y  +  5  ==  0. 

14.  x2  +  9y2  =  9.  15.    4x2  +  9y2  =  1. 

16.  x2  +  4y2  -  2x  -  IGy  +  IG  =  0.  17.    4x2  +  y2  +  24x  +  20  =  0. 

18.  25x2  +  G4y2  +  lOOx  +  128y  +  IGO  =  0.  19.    lGx2  +  25y2  =  400. 

20.  x2  +  4y2  =  12,  21.    5x2  +  9y2  _  20x  -  25  =  0. 

22.    9x2  +  25y2  =  225.  23.    ^  +  ^  =  1. 

3t)       169 

24.  24x2  +  25y2  =  210.  25.  9x2  +  8y2  =  209- 

26.  30x2  _  y2  =r  36.  27.  9y2  -  10x2  =  144. 

28.  9x2  _  64y2  =  4.  29.  4x2  _  y2  _  gx  -  4y  —  4  =  0. 

30.  2x2  —  3y2  -  8x  +  18y  -18  =  0.                   31.    40x2  _  0y2  =  (J40. 

32.  2x2  -  5y2  =  70.  33.  5x2  _  4y2  =  125. 

34.  7x2  -  3y2  +  20  =  0.  35.  25x2  -  9y2  =  IG. 

36.  9y2  -  x2  =  35. 

37.  (1)  (0,0) ;  (2)  y  =  0  ;  (3)  7,  2  ;  _(4)   (±  7,  0) ; 

(5)  f  V5  ;  (6)  (±  3V5,  0);    (7)   x  =  ±  f|  Vs. 

38.  (1)   (0,  0)  ;  (2)   X  =  0 ;  (3)    ^^,  V|;  (4)  (0,  ±  Vi) ; 

(5)   Vi;  (6)  (0,  ±  V^);  (7)  y  =  ±  Vj. 

39.  (1)   (0,  0)  ;  (2)  (0,  -  |)  ;  (3)   x  =  0 ;  (4)  4y  -  7  =  0. 

40.  (1)   (0,  0) ;  (2)   X  =  0  ;  (3)  4,  3 ;  (4)  (0,  ±  3) ;  (5)   ^  ; 

(6)   (0,  ±  5)  ;  (7)  5y  ±  9  =  0  ;  (8)  4y  ±  3x  =  0. 


366  PLANE   ANALYTIC    GEOMETRY. 

41.  (1)  (0,  0)  ;  (2)   (4,  0)  ;  (3)  y  =  0  ;  (4)   X  +  4  =  0. 

42.  (1)  (0,0);  (2)  y  =  0;  (3)   Vs,  V2;  (4)  (±  Vs,  0) ;  (5)    Vf ; 

(6)  (±  V5,  0)  ;  (7)  X  =  ±  I  V5  ;  (8)  y  V3  ±  x  ^  =  0. 

43.  (1)  (0,  0) ;  (2)  (0,  V) ;  (3)  x  =  0 ;  (4)  4y  +  15  =  0. 

44.  (1)  {-  i,  f)  ;  (2)  y  =  f ;  (3)  1,  V2  ;  (4)  {h  f),  (-  f ,  |)  ;  (5)   Vs  ; 

(6)  (-i±V3,f);  (7)  x=-i±iV3;    (8)  y  -  J  =  ±  V2  (  x+ i). 

45.  (1)  (2,  -  1)  ;  (2)  (2,  |) ;  (3)  x  =  2  ;  (4)  4y  +  15  =  0. 

46.  (1)   (1,3);  (2)  x  =  l;  (3)  1,  VS;  (4)   (1,  3±V3); 

(5)    Vf ;  (6)  (1,  3  ±  \^) ;  (7)  2y  =  6  ±  3  V2. 

47.  (1)  (-  h  i) ;  (2)  (-  i|,  1) ;  (3)  2y  -  1  -  0  ;  j4)  24x  -  1  =  0. 

48.  (1)   (-  3,  -  2)  ;  (2)  y  +  2  =  0  ;  (3)   Vl^,  i  V26  ;_ 

(4)  (-3±Vl3^-2);  (5)  iV2;  (6) . (- 3  ±  i  V26,  -2); 

(7)  X  +  3  ±  V26  =  0. 

49.  (1)  (I,  -i);  (2)  2x-3  =  0;j3)  iV2,  iV3;  (4)  (|^-i±iV3); 

(5)  i  VlO  ;  (6)  (3,  -  i  ±  ^  VSO)  ;    (7)  15y  +  5  ±  VSO  =  0 ; 

(8)  6y +  2  =  ±  V6(2x-3). 

50.  (1)   (1,^2);  (2)  y=j-2;  (3)    V?,  VS;  (4)   (1  ±  V?,  -2); 
(5)    V-V°- ;  (6)  (1  ±  VlO,  -  2) ;    (7)  lOx  =  10  ±  7  VTo  ; 

(8)  y +  2  =  ±  Vf  (x-1). 

51.  (1)  (2,-1);  (2)  y  =  -l;  (3)  5,3;  (4)  (7,-1),  (-3,-1); 

(5)  f  ;  (6)  (6,  -  1),  (-  2,  -  1)  ;  (7)  4x  =  33,  4x  +  17  =  0. 

52.  (1)   (i,  -  -\5-) ;  (2)   (i,  -  0)  ;  (3)  2x  =  1 ;  (4)  2y  +  13  =  0. 

53.  (1)  (i,0);  (3)  I;  (5)  0. 

54.  (1)   a,  0);  (2)  y  =  0;  (3)  |,  |;  (4)   (-3,  0),  (4,  0);  (5)    V2; 

(6)  (i  ±  I  V2,  0) ;  (7)  4x  =  2  ±  7  V2  ;  (8)  2y  =  ±  (2x  -  1). 

56.  ^-y'  =  ±^. 

a'^      b^  a 

_,     (x  —  a)2  ,    y-       ,    , 


CHAPTER   VII. 

1.  y  -  1  =  0,  X  +  2  =  0.  2.    3x  +  2y  =  0,  2x  -  3y  =  0. 

3.  X  —  y  +  3  =  0,  x  +  y-3  =  0.         4.    x+y  +  4  =  0,  x-y  +  2=0. 

5.  3x  -  2y  +  3  =  0,  2x  +  3y  -  11  =  0. 

6.  x  -  2y  —  4  =  0,  2x  +  y  -  3  =  0. 

7.  3x  +  y  -  7  =  0,   X  —  3y  -  9  =  0.  8.    x  -  y  +  1  =  0. 


55. 

x2       y2  _       2x  _ 
a2  +  b2  ~  *  a  " 

60. 

-«^>g  =  Mb>c, 

62. 

Ellipse  with  centre  at  origin. 

ANSWERS.  367 

9.  X  -  3y  +  4  ±  V2I  =  0.  10.    X  -  2y  ±  Vl7  =  0. 

11.  2x  —  3y  —  5  =  0,  2x  —  3y  -3  =  0. 

12.  2x  —  y  ±  V39  =  0,  X  —  2y  ±  6  =  0. 

13.  x-2y-3  =  0,  x-2y-2  =  0.  14.    81x  +  54y  -  7  =  0. 
15.  X  -  y  +  6  =  0,  2x  -  2y  -  13  =  0.           16.    8x  +  14y  -  36  -  0. 

17.  3x+4y  +  6  =  0;  (2,  -3),  (f,  -  |). 

18.  4x-3y  +  24  =  0.  19.    x  +  y  -  2  =  0  ;   (3,  -  1). 

20.  7x  +  12y  -  72  =  0,  14x  -  9y  -  23  =  0. 

21.  9x  -  4y  +  2  =  0,  3x  +  4y  +  6  =  0. 

22.  2(5  + V3)x  -  5(2+  V3)y  =21.  24.    12_±2_^  _ 

25.  tan  \v         26.   tan  ^  |,  tan  ^  f  f.         28.    tan  ^^,  tan  '— • 

29.  132x2  +  65y2  -  2340  =  0.  30.    y2  =  2x. 


CHAPTER   VIII. 


1.  3i.  2.    (p(4n -1),  ±2p  V4n  -  1). 

6.  3x  -  4y  +  12  =  0,  4x  +  3y  -  34  =  0. 

7.  X  -  2y  +  6  =  0,  3x  —  2y  +  2  =  0.  8.    x  -  2y  +  9  =  0. 
9-  (i^^a^/^).                          10.    2x-y-12  =  0. 

11.  4x  +  4y-27  =  0.  12.    -l±iV5. 

13.  X  —  y  +  p  =  0,  X  +  y  +  p  =  0  ;  x  +  y  —  3p  =  0,  x  —  y  -  3p  =  0. 

16.  /(xi  +  2p)2_       4p(xi  +  2p)\    4p^(xi  +  p)^ 

\       xi      '  yi        /'         xi 

24.  y-2x  =  0.  25.    fV2.  30.    x^  +  y-^  +  3x  -  6y  =  0. 

31.  x2  +  y2  —  5px  =  0.  32.     Vp  (p  +  Xi). 

35.  36x2  +  36y2  -  15x  -  26y  -  9  =  0.  39.    18y  -  5  =  0. 

40.  y  -  1  =  0  ;  —  |.  41.    i.  42.    3x  +  lOy  +44=0. 

43.  y  +  5  =  0.  44.    5x  +  8y  +  16  =  0. 

48.  y2  =  p{x  — p).  49.    x  +  2p  =  0,  x»  =  py-. 

52.  x2  —  2ay  —  a2  =  0.  55.    Parabola.  58.    Parabola. 

59.  (x  —  y)2  =  4  V2p(x  +  y)  —  8p2,  or,  symmetrically,  x^  +  y-  =  a^. 

CHAPTER    IX. 

1-  Vi,    V|;    Vf  ;    (±  V^,  0);    x  =  ±  V| ;    fVs. 

2.  3x2  +  8y2  =  35.  3.    100x2  +  30y2  =  2025. 
4.  5x2  +  9y2  =  405.                            6.    i;   3x2  +  4y2  =  Sa*. 
7.  iV3  ;  x2  +  4y2  =  a2.  8.    iV2. 


368  PLANE    ANALYTIC    GEOMETRY. 

9.  18x  ±  12  Vl5y  -  29  =  0,  6  Vl5x  q:  9y  -  2  Vl5  =  0. 

10.  7|,  If  11.    x  +  yV3-3  =  0,    3x-yV3-l=0. 

12.  2x  +  .3y  ±  6  V2  =  0.        13.  VSx  -  y  ±  Vl7  =  0. 

14.  (j\\,  j%Oj).  15.    ex  -  y  +  a  =  0,   x  +  ey  +  ae^  =  0. 

17.  (  ±  '  ±    .  ) .  18. '  -  xi. 

\      Va2  +  b2         Va2  +  b2/  xi         a^ 

a         b  \      /       a  b 


^^•(V2'    V2)'    (      V2'        V2) 


22.  a2yix  -  b2xiy  =  0,      ,       ^     '     =  ■  27.    2x  -  3y  =  0. 
Vb*xi2  +  a4yi2 

28.  X  +  3y  =  0,  6x  -  6y  =  0.                29.  x  -  4y  =  0,  8x  +  3y  =  0. 

30.  4x  +  3y  =  0,  3x  -  25y  =  0.            31.  (3  V2,   ^/2),   (-  3  ^A,  -  V2). 

32.  3x  -  2y  -  8  =  0.                               33.  ay  -  bex  =  0,  aey  +  bx  =  0. 

39.  x±2y  =  0.              41.    (—3,   -  Vi),  210°.             42.    (|V3,  1). 

48.  4x2  +  5y2  =  20.       49.    2x2  +  2y2  =  9.  53.    x2  +  y2  =  (a  +  b)2. 


54. 

^  +  J't  =  2.             55.    b2x2  +  a2y2  -  ab^x  =  0. 
a^       b^                                             -' 

56. 

-'+g-^-^  =  0.         58.    e  =  Jl-tan2f,  ifa<^- 
a2       b2        a2         b2                                   \                  2                  2 

CHAPTER   X. 

2. 

52x2 -117y2  =  570.               3. ^^^  + ^^  =  1. 

1  +  V29      6  -  V29 

4. 

10x2  -  5y2  =  19.                     5.    8x2  _  y2  -  ig  or  -  x2  +  8y2  =  124, 

6. 

x2-y2==21.                            7.    3x2-y2  =  3a2. 

11. 

±iV5.                                  12.    cos-^^-^^- 

15.  9x -4y  ±  0V5  =  0,  4V5x +  9V5y  ±78  =  0. 

16.  2x  -  5y  ±  8  =  0.  17.   2|,  H  ;   3J,  2^. 

19.  (ae,  ±  -\  tan-i  (±  e).     26.    4x  -  .3y  =  0,  3x  -  2y  =  0. 

27.  4x-.3y-13  =  0.  28.    (J V3,  V3),  {- J^/3,  -  V3). 

2^-  £-43  =  ^-  ^2.    4xy  =  5. 

43.  a  =  b  =  V2;    e  =  V2;    (1,  1),   (-1,   -1). 

44.  a  =  e,  b  =  2V3,  e  =  2;  (2V3,  —  2V3),   (-2V3,  2V3). 
47.  Straight  line.  48.    Equilateral  hyperbola. 


ANSWERS.  369 

49.    Hyperbola.  50.    Equilateral  hyperbola. 

53.    x2  +  y2  =  a2  —  b^.  54.    x^  +  y2  =  a^. 

CHAPTER   XI. 

1.  Hyperbola,  4x"2  —  y"-^  =  1 ;  centre,  (1,  2) ;  slope  of  axis,  2. 

2.  Parabola,  (  y' ;==  )  — ■p=  (  x' ;=  )  ;  slope  of  axis,  i. 

V  Vio/  VioV        7V10/ 

3.  Imaginary  ellipse. 

4.  Hyperbola,  4x"2  —  9y"2  =  33  ;  centre,  (—  1,  0) ;  slope  of  axis,  1. 

5.  Hyperbola,  2x"2  —  2y"2  =  —  13  5  centre,  (2,  —  ?,)  ;  slope  of  axis,  1. 

6.  Two  coincident  straight  lines. 

7.  Ellipse,  9x"2  +  4y"2  =  36  ;  centre,  (-1,2);  slope  of  axis,  1. 

8.  Two  intersecting  straight  lines.      9.    Two  intersecting  straight  lines. 

10.  Parabola,  (  y'  H 7=  |  = ;=  (  x' '-^  ]  ;  slope  of  axis,  —  1. 

V  8V2/  4V2\  I6V2/ 

11.  Two  parallel  straight  lines. 

12.  Ellipse,  x"2  +  2y"2  =  2  ;  centre,  (2,  —  1) ;  slope  of  axis,  f . 

13.  Point.  20.    2x2  +  Sxy  +  y2  +  12x  -  13y  -  50  =  0. 

21.    x2  +  y2  —  12x  +  4y  +  15  =  0.  22.    x2  -  xy  +  y2  -  a2  =  0. 

23.  (2x  +  y  -  3)  (3x  +  y  -  10)  =  0. 

24.  x2  -  2xy  +  y2  —  22x  +  18y  +  101  =  0, 

529x2  _  5or,xy  +  121y2  -  7126x  +  .3402y  +  23957  =  0. 

26.  Hyperbola,  centre  at  middle  of  base. 

27.  Hyperbola,  centre  at  middle  of  base. 

28.  Hyperbola,  asymptotes  parallel  to  coordinate  axes. 


PART   II. 
CHAPTER   I. 


1.    Vl49.                2.    (3,  4,  -0),  (4,  5,  -8).  4.  (4,  -  17,  21). 

5.    (5,  -  6,  5).       6.    (3,  -  2,  1),  (-  |,  |,  V)-  7.  Vl4  ;(-!,-  1,  8). 

8.    iV206;  (i^s-ff,  il).        9-    cos  '  Vi.  10.  — -i  — .±  — • 

11.    -|=,  -4-,  -7=-           12.    7t.  13.  I,  |V3,  f. 
V29         V29    V29 

14.    (13,  cos"'  xV.  tan~'  4). 


3T0 


SOLID    ANALYTIC    GEOMETRY. 


CHAPTER   IL 


4.  x2  +  y2  ±  2ay  =  0. 

14.  X  -  3  =  0. 

16.  y*  =  16p2  (x2  +  z2). 

17.  (a)  X  -  a  =  0  ;  (b)  y2  +  z2  =  a2 ;  (c)  y2  +  z2  -  (mx  +  b)2  =  0 


5.    y2  +  z2  ±  2ay  ±  2az  +  a2  =  0. 
15.    m2x2  —  y2  —  z2  =  0. 


-1 


CHAPTER   III. 

1.    X  +  6y  +  5z  =  ±  5  V62.  2 

3.    7x  +  4y  -  5z  =  0  ;  0.  4.    cos' 

6.    X  —  l^-'     _    =z  — 1.  7.    — 7r-=- 

V2  3 

3       -2       3 
yj22    V22    \^ 
10.    3x  -  2y  +  3z  -  21  =  0.  11. 

12.    X  +  y  +  z  =  0. 


5.    9. 

z  +  1 


-2  3 

9.    2x  +  y  -  7  =  0,  z  +  3  =  0. 
17  26  11 


'1086 


13        /2  8       1_1       6  4\ 


)86       VIO86 

14.   sin~^ 


V671 


15.    X  +  y  +  z  +  3  +  3  V3  =  0.  16.    4y  +  9  =  0  ;  2x  +  6z  -  1  =  0. 

17.    3x  -  4y  +  12z  +  102  =  0  ;  f|.  18.    9x  -  4y  -  5z  -  16  =  0. 


19. 


1 _y-3 _  z  +  5 
~      6      ~ 


2  6  7 

23.    (-1,  -2,  -3). 


21.   cos 


-1     9 


2V2r 

24.    6x-18y-77z-29  =  0. 
25.    2x  +  4y  +  17z  +  1  =  0.  26.    2x  +  7y  -  9z  +  14  =  0. 

27.    2x  +  6y  —  3z  +  7  =  0,  12x  -  3y  +  2z  +  43  =  0. 

29.  (1  +  I V2,  f,  V).  30.    (V,  -  tV.  0). 

31.  (-\^  -  f ,  -  -1/).  32.    (-  1,  -  2,  -  3). 

33.  VlO.  34.    1. 

38.  Circle.  39.   Plane,  sphere. 

44.  X  +  3y  +  z  -  1  =  0,  X  +  y  +  2  =  0. 


CHAPTER   IV. 

2.    3x2  +  3y2  -  z2  =  0.  3.    2x2  +  3y2  -  6z2  =  0. 

4.    2x2  +  3y2  +  3z2  =  1 ;    2x2  +  3y2  +  2z2  =  1. 


14. 


^±^  =  2p,z  +  ,,. 


24.    xix  +  yiy  +  ziz  =  a^. 


ANSWERS.  371 

„„     /-a2A     -a2B     -a2C\  ^„     /-a2A     -  b2B     -c2C\ 

34.  X  cos  a  +  y  cos  /3  +  z  cos  7  =  0. 

^^     xcosa   ,   ycosiS      „  ^ 

35.    r— ±  =^-7— -^  — 2pcos7  =  0. 

di^  b- 

a2A  b2B  -c2C 

36 


Va4A2  +  b4B2  +  c*C2    Va''A2+  b^B^+ciC^      Va''A2  +  b*B2  +  0*02. 

49.    (  ±  —^  .  ±  — ='  0)- 
V        V13  V13       / 


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